Winglish Coaching Centre, Puduvayal 43 XI Mathematics Made Easy 3.1 BASIC RESULTS OF TRIGONOMETRY I. Angles AOB is formed by two rays OA and OB sharing common point O called vertex of angle. If we rotate ray OA about its vertex O and takes position OB, OA and OB are called initial side and terminal side of the angle produced. An anticlockwise rotation generates a positive angle (positive sign), clockwise rotation generates negative angle (negative sign). One full anticlockwise or clockwise rotation of OA back to itself is one complete rotation II. Different Systems of measurement of angle i) Sexagesimal system Right angle is divided into 90 equal Degrees. Degree is divided into 60 equal Minutes Each minute into 60 equal Seconds. 1, 1’ and 1” denote a degree, a minute, a second ii) Centesimal system Right angle is divided into 100 equal Grades Each grade is subdivided into 100 Minutes Each minute is subdivided into 100 Seconds. 1 g denotes a grade. iii) Circular system Radian measure of an angle is introduced using arc lengths in a circle of radius r. 1 c denotes 1 radian measure. III. Degree Measure Unit of angle measurement represented by degreeOne complete rotation is split into 360 equal parts and each part is one degree, denoted by 1. 1is 1/360 of one complete rotation. To measure a fraction of angle minutes and seconds are introduced. One minute 1corresponds to 1/60 of a degree A second (1’) corresponds to 1/60 of a minute (or) 1/3600 of a degree. IV. Classification of a pair of Angles i) congruent angles Two angles that have the exact same measure ii) complementary angles Two angles having their measures adding to 90 0 iii) supplementary angles. Two angles that have measures adding to 180 0 iv) conjugate Angles Two angles between 0and 360are conjugate if their sum equals 360. V. Angles in Standard Position An angle is in standard position if its vertex is at origin and initial side is along the positive x-axis. An angle is in first quadrant, if in standard position, terminal side falls in the first quadrant. Angles in standard position having terminal sides along x-axis or y-axis are quadrantal angles. 0, 90, 180, 270and 360are quadrantal angles. Degree of quadrantal angle is multiple of 90. VI. Coterminal angles One complete rotation of a ray in anticlockwise direction results in an angle measuring of 360. By continuing the anticlockwise rotation, angles larger than 360can be produced. If we rotate in clockwise direction, negative angles are produced. Angles in standard position that have the same terminal sides are coterminal angles . If and β are coterminal angles, β = + k(360), k is an integer. Coterminal angles differ by integral multiple of 360
Transcript
Winglish Coaching Centre, Puduvayal 43 XI Mathematics Made Easy
One complete rotation is split into 360 equal parts
and each part is one degree, denoted by 1.
1 is 1/360 of one complete rotation.
To measure a fraction of angle minutes and
seconds are introduced.
One minute 1 corresponds to 1/60 of a degree
A second (1’) corresponds to 1/60 of a minute (or)
1/3600 of a degree.
IV. Classification of a pair of Angles
i) congruent angles
Two angles that have the exact same measure
ii) complementary angles
Two angles having their measures adding to 900
iii) supplementary angles.
Two angles that have measures adding to 1800
iv) conjugate Angles
Two angles between 0 and 360 are conjugate if
their sum equals 360.
V. Angles in Standard Position
An angle is in standard position if its vertex is at
origin and initial side is along the positive x-axis.
An angle is in first quadrant, if in standard
position, terminal side falls in the first quadrant.
Angles in standard position having terminal sides
along x-axis or y-axis are quadrantal angles.
0, 90, 180, 270 and 360 are quadrantal angles.
Degree of quadrantal angle is multiple of 90.
VI. Coterminal angles
One complete rotation of a ray in anticlockwise
direction results in an angle measuring of 360.
By continuing the anticlockwise rotation, angles
larger than 360 can be produced.
If we rotate in clockwise direction, negative
angles are produced.
Angles in standard position that have the same terminal sides are coterminal angles .
If and β are coterminal angles, β = + k(360), k
is an integer.
Coterminal angles differ by integral multiple of
360
Winglish Coaching Centre, Puduvayal 44 XI Mathematics Made Easy
1. Express 59.0854into degrees, minutes, seconds
59.0854 = 59 + 0.0854
= 59 +.0854 o1
60'
= 59 + 5.124’
= 59 +5’ + 0.124’
= 59 +5’ +0.124’ 1'
60"
59.0854 = 595’7.44”
2. Express Complementary of 3618’47”into degrees, minutes, seconds
Complementary = 90 − 3618’47”
= 5341’13”
3. Express Supplementary of 34.8597into degrees, minutes, seconds
34.8597 = 3451’35”
Supplementary = 180 –3451’35”
= 1458’25”
4. Identify the quadrant in which given angle lies
(i) 25 (i) first quadrant
(ii) 825 (ii) second quadrant
(iii) −55 (iii) fourth quadrant
(iv) 328 (iv) fourth quadrant
(v) −230 (v) second quadrant
5. Check whether 417 and −303 are coterminal. If so find another coterminal angle
If , β coterminal angles, β - = k(360)
417− (−303) = 720
= 2 (360)
417 and −303 are coterminal
57, 417 and −303 have same initial side and terminal side but with different amount of rotations, so they are coterminal angles.
6. Write coterminal angles pairs for 30, 280, -85
30, 390 280, 1000
−85, 275
7. Find a coterminal angle with measure of θ such
that 0 ≤ θ < 360
(i) 395 35
(ii) 525 165
(iii) 1150 70
(iv) −270 90
(v) −450 270
I. Radian measure of an angle
Ratio of arc length it subtends, to radius of the circle in
which it is the central angle.
rs r
sθ
Hence,
radian
radius
length arc centre. at subtended Angle
slength arc Its
r radius of circle a Consider
II. Properties of Radian measure
i) All circles are similar.
For a given central angle in any circle, the ratio of the
intercepted arc length to the radius is always constant.
ii) When s = r, θ = 1 radian.
1 radian is the angle made at the centre of a circle by
an arc with length equal to radius of the circle.
iii) s and r have same unit
θ is unitless and no notation to denote radians.
iv) θ = k radian measure, if s = kr.
θ = 1 radian measure, if s = r
θ = 2 radian measure, if s = 2r
Radian measure tells us how many radius lengths,
need to sweep out along circle to subtend the angle θ.
v) Radian angle measurement can be related to the
edge of the unit circle.
we measure an angle by measuring the distance
travelled along the edge of the unit circle to where the
terminal side of the angle intercepts the unit circle .
III. Difference b/w Degree and Radian Measures
In measuring temperature, Celsius unit is better
than Fahrenheit as Celsius was defined using 0
and 100 for freezing and boiling points of water.
Radian measure is better for conversion and
calculations.
Radian measure is convenient for analysis;
Degree measure is convenient to communicate
the concept between people.
Scale used in radians is much smaller than the
scale in degrees. Smaller scale makes graphs of
trigonometric functions more visible and usable.
Winglish Coaching Centre, Puduvayal 45 XI Mathematics Made Easy
IV. To convert radians into degrees or degrees into
radians
In unit circle, a full rotation corresponds to 360
or 2π radians, the circumference of the unit circle.
radiansradians
radianradian
radians
1801801
1801801
180
3602
xx.
x x
π radians
oo
oo
o
o
V. Properties of π
i) The ratio of the circumference of any circle to its
diameter is always a constant.
This constant is denoted by irrational number π.
ii) Values of π and 7
22 correct to four decimal places
are 3.1416 and 3.1429
π and 7
22 are approximately equal correct upto
two decimal places.
Hence, π ≈7
22
iii) 1 radian ≈ 57O 17’45”
1o ≈ 0.017453 radian
1’ ≈
60180
≈ 0.000291 radian.
1” ≈
6060180
≈ 0.000005 radian.
VI. Radian measures and corresponding degree
measures for some known angles
VII. Area of the sector
360r2Perimeter
radian in sector of Area
degreein sector of Area
2
2
360
2
2
2
r
r
r
Solved Problems
1. Express following angles in radian measure:
Example : i) 18 ii) −108
radians
radians
radians ii)
radians
radians
radiansi)
5
3180
108108
1801
10
180
1818
1801
oo
o
o
o ..
i) 30 radians6
ii) 135 radians
4
3
iii) −205 radians36
41 iv) 150 radians
6
5
v) 330 radians6
11
2. Find degree measure of given radian measures
o
o
o
200radiansvii)radiansvi)
radiansv)radiansiv)
radiansiii)180
radians ii)
7
180
180radians6 i)
180 radians
180 radians
9
10420
3
7
725
220
9
306
3655
11343
6
6
1
722
o
oo
oo
o
o
o
o
3. What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?
meters 31.82
1km nceCircumfere
r path circular of Radius
2522
71000
100025
5
r
mr
4. In a circle of diameter 40 cm, a chord is of length 20 cm. Find the length of the minor arc of chord.
traiangle lequilatera make chord of Ends
cm 20.95
arc minor ofLength
60 Angle Central
20cm circle of Radiuso
3
20
1802060
0
o
r
Winglish Coaching Centre, Puduvayal 46 XI Mathematics Made Easy
5. Find the length of an arc of a circle of radius 5 cm subtending a central angle measuring 15
cm
5
arc the ofLength
angle central
cm circle radiusof
12
18015
18015
15
5
rs
ro
6. What is the length of the arc intercepted by
central angle 41 in a circle of radius 10 ft?
7.16feet
f18
1
10
arc the ofLength
angle central
circle radiusof
t
rs
ftr
o
o
7
2241
18041
41
10
7. Find the degree measure of the angle subtended at the centre of circle of radius 100 cm by an arc of length 22 cm.
'
'.
).(
cm arc the ofLength
angle central
cm circle radiusof
3612
606012
612
22
7
5
922
22180
100
22
100
o
o
o
s
r
8. If arcs of same lengths in two circles subtend
central angles 30, 80, find ratio of their radii.
38
30
80180
80180
30
21
2
1
21
2211
2
2
21
21
::
Given)(
circlesgiven anglestwo central,
arc. the oflength ,
circlesgiven two the of radii
1
rr
r
r
rr
rr
ll
ll
,rr
1
9. If in two circles, arcs of the same length subtend
angles 60,75 at centre, find ratio of their radii.
45
7560
21
21
::
rr
rr oo
10. The perimeter of a sector of a circle is equal to the length of the arc of a semi-circle having the same radius. Express the angle of the sector in degrees, minutes and seconds.
27'16"65 sector of Angle
radian1.1416
2-3.1416
semicircle of arc ofLength Perimeter
leofsemicirc angle Central
sector of angle Central
semicircle and sector Radius r
o
2
2 rrr
11. An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on the edge of propeller will rotate in 1 second.
sec/
secdeg/360
mindeg/360 rotation
deg360 rotation 1
o
o
o
06000
60
1000
10001000
12. A train is moving on a circular track of 1500 m radius at the rate of 66 km/hr. What angle will it turn in 20 seconds?
o
rs
r
14
1500
203318
6060
radian.
radian
2018.3320sin covered Distance
1000661sin covered Distance
km/hr 66 speed
1500m Radius
13. Circular metallic plate of radius 8 cm thickness 6 mm is melted and molded into a pie of radius 16 cm thickness 4 mm. Find angle of sector
circular metallic plate with thickness is a cylinder
4
3
135
401616
3606088
3602
o
HRh
H
R
h
r
.
.
r
height Areaheight Area
Cylinder of Volume sector of Volume
0.6cm height
8cm Radius
0.4cm height
16cm Radius
2
: Cylinder: Sector
Winglish Coaching Centre, Puduvayal 47 XI Mathematics Made Easy
BASIC TRIGONOMETRIC RATIOS & IDENTITIES
1. Prove that
cos
sin 1
1 sec - tan
1 - sec tan
cos
sin
sectansectan
]sec)[tansec(tansectan
)]tan(sec)[sec(tansectan
)tan)(sectan(sec)sec(tansectan
)tan(secsectanLHS
1
1
11
11
1
22
2. Prove that (secA−cosecA)(1+tanA+cotA)=tanAsecA−cotAcosecA
AecAAA
ecAecAecA
AecA
AecAA
A
ecAecA
A
ecAA
ecA
AecA
ecA
AecAA
A ecAAA
AecAAecAA
A) AecA)(A(
cotcostansecsec
coscoscos
seccos
secseccossec
sec
coscos
sec
cossec
cos
seccos
cos
secseccossec
cotcoscotsec
tancostanseccossec
cottancossec
1
3. Eliminate θ from a cos θ = b and c sin θ = d,
222222
22
2222
2
2
2
222
2
22
2
22
1
21
21
cbdaca
ca
cbda
a
b
c
d
c
d
d θc
a
b
ba
sincos),()(
)...(sin
sin
)....(cos
cos
4. If a cos θ − b sin θ = c, show that
.cba ba 222 cossin
.cba ba
cbaba
ca
b
cba
ab ba
ab
ab ba
abcba
c b a
222
2222
2222
222
22222
22222
22222
22222
22
2
12
)cossin(
)cossin(
)sin(cos
)sin(cos
sincos
sincos)cossin(
(1)Using
cossin
sincos)cossin(
)....(cossinsincos
)sincos(
5. If sin θ + cos θ = m, show that
4
134 2266 ) - (m -
sincos where m2 ≤ 2.
4
134
14
131
3
14
1
12
2
22
22
2222
22
323266
2222
2
222
22
) - (m -
)()(m
θ)θθ(θ
θ)θ(
θ)(θ)(θθ
).....()(m
θθ
mθθ
mθθθθ
mθ)θ(
mθθ
using
cossincossin
sincos
sincossincos Now,
cossin
cossin
cossincossin
cossin
cossin Let,
6. If
sincos
sin
1
2 y prove y
sin
sincos
1
1
sin
sincossinsin
)sincos(sin
)sin(sinsin
)sincos(sin
cos)sin(
)sincos(sin
)cossin)(cossin(
)sincos(sin
)sincos)(sincos(
)sincos(sin
)sincos( by multiplysincos
sincos
sincos
sin
1
122
12
121
12
1
12
11
12
11
12
11
1
1
2
2
22
22
y
7. If 1
sin
sin
cos
cos2
4
2
4
prove that
sin
sin
cos
cosii)sinsinsinsini)
2
4
2
42222 2
sin
sin)cosc()cos(cos
cos
cos
sin
sin)cos c()cos(cos
cos
cos
sin
sin)sin sin()cos(cos
cos
cos
sin
sinsincos
cos
cos
sincos sin
sin
cos
cos Let
2
222
2
2
2
222
2
2
2
222
2
2
2
4
2
4
2
4
2
4
22
22
22
22
22
11
os
os
Winglish Coaching Centre, Puduvayal 48 XI Mathematics Made Easy
).....(sinsin
sin sin sinsin sin sin
)sin( sin)sin( sin
cos sin cos sin
sin
sin
cos
cos
)........(coscos
coscos
sin
sin
cos
cos)cos(cos
222222
2222
2222
2
2
2
2
2
2
2
2
22
2
11
1
0
0
22
22
2
2
1
2 22
2222
222244
sin
sin
cos
cos
sin
sin
cos
cos
sin
sin
cos
cosii)
sinsin
sinsinsinsin
sinsinsinsinsinsini)
2
4
2
4
2
4
2
4
2
4
2
4
8. Show that
./k ec
2323 2 costansec if 22 1 ktan
./k
ec
k
k
k
k
232
3
2
2
2
3
33
2
22
22
22
2
1
1
2
2
111
1
)(sec
)(secsec
)tan(sec
sectansec
sin.
cos
sinseccostansec
sec
sec
tan
tan
9. If 2
0 and
0
2
0
2
n
n
n
n θ,yθx sincos
0
22
n
nn θθz sincos Prove that zy xxyz
2222
222
22
32
2
0
2
1
111
1
11
1
1
1
1
11
1
cossinsincos
sincoscos
sincos
. . .
G.P infinitean is...coscos
cos
zy x
zy
xx
xxxx
x
θxn
n
xyz
)cossin(cossin
)cossin(cossin
sincoscossin
cossinsincos
sincos
2222
2222
2222
2222
22
1
1
1
1
1
1
10. If sec θ + tanθ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
p
p
p
p
p
p
p
p
p
p
p
ppp
p
p
p
p
pp
p p
2
1
2
1
1
1
1
1
1
1
1
412
1
41
11
2
1
1
1
2
2
2
2
2
2
22
22
22
224
22
2
2
2
cos
sintan
)(
)(
)(
)(
secsin
sec
sec2 Adding By
tansectansec
11. If cot θ (1 + sinθ) = 4m and cot θ (1 − sin θ) = 4n, then prove that
mn)- n(m
mn)(
mn)()(
)- n(m
nm
nm
n
n -
n ) -(
m
m
m ) (
222
22
22222
22
22
222
222
16
1
1611
31
516
4
16442
4162
4
341
2162
4
141
sincot
sincotsincot
:Gives)()(Now,
)...(coscot
coscot)(
)(coscot),()(
)...(coscotcoscot
coscot
)..(sincot
)...(coscotcoscot
: sidesboth Squaring
coscot
)......(sincot
Winglish Coaching Centre, Puduvayal 49 XI Mathematics Made Easy
12. If 3 a ec sincos and 3 b cossec ,
then prove that 12222 b aba )(
23
22
23
22
3
23
3
22
32
32
3
3
3
2
1
1
ba
a
a
a
a
a ec
cos
sin Thus,
sin
cos
sin
cos, to powerRaising
sin
cos
sin
sin
sinsin
sincos
1
22
3
2
3
223
2
3
22
3
223
22
3
2
3
23
22
3
23
22
3
223
22
3
2
3
223
223
222
2222
sincos
)(sincos)(coscos
cos
sin
sin
cos)sin(cos
)sin(coscos
sin)sin(cos
sin
cos
cos
sin
sin
cos)sin(cos
cos
sin
sin
cos
cos
sin
sin
cos)(
b aba
13. Eliminate θ from the equations a secθ− ctan θ= b
and b sec θ + d tan θ = c.
22222
22222
2222
22222
22
22
1
1
0
0
)()()(
)()()(
tansec
gingsubtractin andSquaring
tan,sec
tansec
tansec
tansec
tion multiplica crossUsing
tansec
tansec
2
bacbcadbdc
bacbacbcad
bcad
acb
bcad
bdc
bcad
acb
bcad
bdc
bcad
acb
bcad
bdc
bcadacbbdc
θ
cdb
bc-θa
cdb
b c-θa
TRIGONOMETRIC FUNCTIONS
I. Trig. Functions in Cartesian coordinates
The trigonometric ratios to any angle in terms of radian
measure are called trigonometric functions.
Let P(x, y) be a point other than the origin on the
terminal side of an angle θ in standard position .
x
r
y
rec
y
x
x
yr
x
r
y
y x
r. OP
seccos
cottan
cos sin
r
Let
22
i) Since |x| ≤ r, |y| ≤ r,
|sin θ| ≤ 1 and |cos θ| ≤ 1.
ii) Trigonometric functions have positive or negative
values depending on the quadrant in which the
point P(x, y) on the terminal side of θ lies.
iv) Trigonometric functions is independent of the
points on the terminal side of the angle.
II. Trigonometric ratios of Quadrantal angles
Angle in its standard position for which terminal side
coincides with one of the axes, is a Quadrantal angle.
Consider the unit circle x2+y2 = 1.
Let P(x, y) be a point on unit circle where terminal
side of θ intersects the unit circle.
P) of coordinate-( sin
P) of coordinate-( cos
yyy
xxx
1
1
Coordinates of any point P(x, y)
P(cos θ, sin θ).
III. Values of Trig. functions of quadrantal angles.
Winglish Coaching Centre, Puduvayal 50 XI Mathematics Made Easy
i) x and y coordinates of all points on the unit circle
lie between −1 and 1.
Hence, −1 ≤ cos θ ≤ 1, −1 ≤ sin θ ≤ 1, no matter
whatever be the value of θ.
ii) When θ = 360, terminal side coincides with
positive x- axis.
Hence, sine has equal values at 0 and at 360.
Other trigonometric functions follow it.
iii) If two angles differ by an integral multiple of 360
or 2π, then each trigonometric function will have
equal values at both angles.
IV. Geometrical generalization
tan θ is not defined when cos θ = 0
So,tan θ is not defined when θ = (2n +1)π/2,nZ
V. Trigonometric Functions of real numbers
Consider a unit circle with centre at the origin.
Let the angle zero (in radian) be associated with
the point A(1, 0) on the unit circle.
Draw a tangent to unit circle at the point A(1, 0).
Let t be a real number such that t is y- coordinate
of a point on the tangent line.
For each real number t, identify a point B(x, y) on
unit circle such that the arc length AB is equal to t.
If t is positive, choose B(x, y) in anti clockwise
direction, or choose in clockwise direction.
Let θ be the angle subtended by arc AB at centre.
A function w(t) associating a real number t to a
point on the unit circle.
Such a function is called a wrapping function .
Using sin t and cos t, other trigonometric
functions is defined as functions of real numbers.
Wrapping function w(t) is analogous to wrapping
a line around a circle.
The value of a trigonometric function of a real
number t is its value at the angle t radians.
s = rθ
Arc length t = θ.
sin t = sinθ
cos t = cos θ.
sin t = sinθ = y
cos t = cos θ = x.
B(x, y) = B(cos t, sin t) is a point on the unit circle.
−1 ≤ cos t ≤ 1 , −1 ≤ sin t ≤ 1 for any real number t.
VI. Signs of Trigonometric functions
Consider a unit circle with centre at the origin.
Let θ be in standard position.
P(x, y) be point on unit circle corresponding to θ.
values of x and y are positive or negative depending
on the quadrant in which P lies.
cos θ = x,
sin θ = y
tan θ = y/x
First quadrant:
cos θ = x > 0 (positive)
sinθ = y > 0 (positive)
cos θ and sin θ and all
trigonometric
functions are positive
Second quadrant:
cos θ = x < 0 (negative)
sinθ = y > 0 (positive)
sin θ and cosec θ are positive and others are negative.
Third quadrant:
cos θ = x < 0 (negative)
sinθ = y < 0 (positive)
tan θ and cot θ are positive and others are negative
Fourth quadrant:
cos θ = x > 0 (negative)
Winglish Coaching Centre, Puduvayal 51 XI Mathematics Made Easy
sinθ = y < 0 (positive)
cos θ and sec θ are positive and others are negative
If sin θ and cos θ are known, then the reciprocal identities
and quotient identities can be used to find the other four
trigonometric values.
Pythagorean identities are used to find trigonometric
values when one trigonometric value and quadrant known.
VII. Allied Angles
Two angles are said to be allied if their sum or
difference is a multiple of π/2radians.
Any two angles of θ such as,
....,,,
2
3
2
are all allied angles .
VIII. Trigonometric ratios of −θ in terms of θ
Let AOL = θ and AOM = −θ.
Let P(a, b) be a point on OL.
Choose a point P’ on OM
such that OP = OP’.
Draw PN rOA intersecting
OM at P’.
AOP=AOP’
PON= P’ON and PON , P’ON are congruent.
Thus, PN = P’N and point P’ is P’(a,−b)
By definition of trig. functions
cot- )cot(- , sec )sec(-
, cosec- )cosec(- ,tan - )tan(-Thus,
cos)cos('
')cos(
cos
sin)sin('
')sin(
sin
OP
aOP
aOP
a
OP
bOP
bOP
b
−θ is same as θ but it is on other side of x- axis.
Flipping (x, y) to other side of x- axis makes it into (x,−y)
So y-coordinate is negated and sine is negated
x-coordinate is same and cosine is unchanged.
Thus, sin(−θ) = −sin θ, and cos(−θ) = cosθ.
IX. Trigonometric ratios of (90 − θ), 0 < θ <π2
sin(90 − θ) = cosθ,
cos(90 − θ) = sinθ,
tan(90 − θ) = cot θ
cosec(90 − θ) = sec θ,
sec(90 − θ) = cosec θ,
cot(90 − θ) = tanθ.
X. Trig.ratios of of the form (90 + θ), 0 < θ <π/2
Let AOL = θ and AOR = (90 + θ).
Let P(a, b) be a point on OL
Choose a point P’ on OR
such that OP = OP’.
Draw rs PM and P’N
from P and P’ on Ox and
Ox’
AOP’ = 90 + θ.
OPM and P’ON are congruent.
ON = MP nd NP’ = OM
Cordinates of P and P’ are P(a, b) and ’(−b, a),
sin
'
coordinate )ocos(90
cos
'
coordinate )osin(90
OP
b
OP
x
OP
a
OP
y
tan(90 + θ) = −cot θ cosec (90+θ) = sec θ
sec(90 + θ) = −cosec θ cot(90 + θ) = −tan θ.
XI. Trig.function of other allied angles
Winglish Coaching Centre, Puduvayal 52 XI Mathematics Made Easy
XII. Summary of Trig.function of allied angles
i) Allied angles of the form 2n2
θ, n Z
That is, −θ, π θ, 2πθ
Form of trigonometric ratio is unaltered
i.e., sine remains sine, cosine remains cosine etc.
ii) Allied angles of the form (2n+1) 2
θ, n Z
2
θ θ,3
2
θθ
Form of trig. ratio altered to its complementary
Add “co” if it is absent
Remove “co” if it is already present
i.e., sine becomes cosine, cosine become sine etc.
iii) For determining sign,
Find out quadrant
Attach appropriate sign(+or−) according to ASTC
iv) Characteristics of Trig. Functions
i) Sine and cosine functions are complementary
sin (90 − θ) = cos θ and cos (90 − θ) = sinθ.
ii) cos θ and sin θ are satisfy inequalities
−1 ≤ cos θ ≤ 1 and −1 ≤ sin θ ≤ 1.
cos θ, sin θ [−1, 1]
XIII. Periodicity of Trigonometric Functions
A function f is said to be a periodic function with
period p, if there exists a smallest positive number
p such that f(x + p) = f(x) for all x in the domain.
Z. n sinx, )2n (xsin .
sinx
. . . )6 sin(x ) 4 sin(x )2 (xsin
ei
sin x , cos x,cosecx,sec x are periodic functions
with period 2π.
tan x,cot x are periodic functions with period π.
XIV. Graph and Important Datas of T- Ratios
Winglish Coaching Centre, Puduvayal 53 XI Mathematics Made Easy
XV. Odd and Even trigonometric functions
Real valued function f(x) is even function if it
satisfies f(−x) = f(x) for all real number x
A real valued function f(x)odd function if it
satisfies f(−x) = −f(x) for all real number x.
As, cos(−x) = cosx for all x, cos x is even function
Also sec x is even function
sin(−x) = −sin x for all x , sin x is odd function.
tan x, cosec x and cot x are odd functions.
f(t) = t − cos t is neither even nor odd function
A function is an even function if its graph is
unchanged under reflection about the y-axis.
A function is odd if its graph is symmetric about
the origin
1. The terminal side of an angle θ in standard
position passes through the point (3,−4).Find the
six trigonometric function values at an angle θ.
Let B(x, y) = B(3,−4), OA be the initial side
OB = terminal side of θ in standard position.
AOB is the angle θ
θ lies in the IV quadrant.
on so andtan
,cos,sin
)(
3
4
5
3
5
4
5
43 22
22
x
yr
x
r
y
yxr
2.
7
62
7
5, is a point on the terminal side of an
angle θ in standard position. Determine the
trigonometric function values of angle θ.
on so andtan
,cos,sin
quadrant. I thein lies
side initial the be OA ,,B = ) ,B( Let
5
62
5
7
7
62
7
5
7
62
149
24
49
25
7
62
7
5
7
62
7
5
22
r
yx
3. If 5
3θsin and angle θ is in second quadrant,
then find other trigonometric functions.
etcecθ
θ
θθ
θθ
3
5
5
3
5
cos,tan,5
4- cos Thus,
quadrant 2in negative cos
4
25
91
sin 1 cos
1 cos sin
nd
2
22
4. Find values of other trigonometric functions
i) 2
1- cos , θ lies in the III quadrant.
Winglish Coaching Centre, Puduvayal 54 XI Mathematics Made Easy
on so and,tan,3
- sin Thus,
quadrant 3in negative sin
3
4
11
cos 1 sin
1 cos sin
nd
2
22
32
2
θ
θ
θθ
θθ
ii) cos θ =2/3, θ lies in the I quadrant.
on so and,tan,5
sin Thus,
quadrant stIin ve sin
5
9
41 sin
2
3
3
3
θ
θ
θ
iii) sin θ = −2/3, θ lies in the IV quadrant.
on so and,tan,5
cos Thus,
quadrantth IVin ve cos
5
9
41 cos
2
3
3
3
θ
θ
θ
iv) tan θ = −2, θ lies in the II quadrant.
on so and1
-
costansin
,1
- cos Thus,
quadrant 2in negative cos
cos
41
tan1 sec
1 tan sec
nd
2
22
5
22
5
5
5
1
5
θ
θ
θθ
θθ
v) sec θ =13/5, θ lies in the IV quadrant.
on so and,tan,12-
sin Thus,
quadrantth IVin ve sin
12
1 sin
13
5cos
5
12
13
13
169
25
θ
θ
θ
5. Find the values of
i) sin(−45) ii) cos(−45) iii) cot(−45)
-1
)cot(45 )cot(-45
1
)cos(45 )cos(-45
1
)sin(45- )sin(-45
2
2
6. Find the value of
i) sin 150 ii) cos 135 iii) tan 120.
3-
)tan(60-
)60 -(180tan 120tan
2
1
)cos(45-
) 45-(180 cos 135 cos
)sin(30
)30-(180sin 150sin
cot
)30 (90 tan120tan
2
1-
)sin(45-
) 45 cos(90 135 cos
)cos(60
)60 sin(90 150sin
o
ooo
o
ooo
o
ooo
o
ooo
o
ooo
o
ooo
2
1
3
30
2
1
θ)(180 Usingθ)(90 Using
7. Find the value of:
iii) cos(300) i) sin(480) v) cot(660)
i) sin 765 iv) tan(1050) ii)sin(−1110)
ii) cosec (−1410)
iii)
4
15-cot vi)
3
19tan vii)
3
11sin
cos 300 = cos(360 -60)
= cos60
= 2
1
sin 480 = sin(360 +120)
= sin120
= sin(180- 60)
= 2
3
sin 765 = sin(2 360 + 45)
= sin45
Winglish Coaching Centre, Puduvayal 55 XI Mathematics Made Easy
= 2
1
cot 660 = cot (2360 -60)
= -cot 60
= - 3
1
tan (1050) = tan (3360 - 30)
= -tan 30
= - 3
1
sin (−1110) = −sin (1110)
= −sin (3360 + 30)
= - sin 30
= 2
1
cosec(−1410) = −cosec (1410)
= −cosec (4360 − 30)
= cosec 30
= 2
4
15-cot =
4-4cot-
=
4cot
3
19tan =
3
6tan
=
3
tan
= 3
3
11sin =
3
4sin
=
3
sin
= 2
3
8. Prove: tan315cot(−405)+cot 495tan(−585)= 2
= tan(360−45)[−cot(360+45)]+cot(360+135)
[−tan (360 + 225)]
= [−tan 45] [−cot 45] + [−tan 45] [−tan 45]
= (−1)(−1) + (−1)(−1)
= 2.
9. Prove that
cotcos))cosec(360tan(- )sin(270
)cos(- )sin(90 )cot(180 2
cotcos
tanot
os cot
)cosec(-tan )os(-
cos os cot
))cosec(360tan(- )sin(270
)cos(- )sin(90 )cot(180
2
2
c
c
c
c
10. Find all the angles between 0 and 360 which
satisfy the equation sin2 θ =34
o
o
1202
3
602
3
2
3
4
,sinwhen
,sinwhen
sin
3 sin 2
11. Prove 2 9
sin18
sin 9
sin 18
sin 2222
47
211918
918
2
cos 9
sin cos 18
sin
2 sin
9 sin
2 sin
18 sin
222
2222
12. Determine whether the functions are even, odd
or neither i) sin2 x − 2 cos2 x − cos x ii) sin (cos(x))
iii) cos (sin(x)) iv) sin x + cos x
i) Let f(x) = sin2 x − 2 cos2 x − cos x
sin(−x) = −sin x and cos(−x) = cos x
f(−x) = f(x)
Thus, f(x) is even.
ii) Let f(x) = sin (cos(x))
f(−x) = f(x)
f(x) is an even function.
iii) f(x) = cos (sin(x)) ,
f(−x) = f(x),
Thus, f(x) is an even function.
iv) f(x) = sinx + cos x
f(−x) f(x) and f(−x) ) − f(x)
Thus, f(x) is neither even nor odd.
Winglish Coaching Centre, Puduvayal 56 XI Mathematics Made Easy
Trigonometric Identities Compound Angles formulas
Compound angles are algebraic sum of two or more angles.
Trigonometric functions do not satisfy the functional
relations like f(x + y) = f(x) + f(y) and f(kx) = kf(x), k is
a real number.
cos(+β) cos +cos β
sin(2) 2sin
tan 3 3tan, . . ..
1. Prove that cos( + β) = cos cos β − sin sin β
Consider the unit circle with centre at O.
Let P = P(1, 0).
Let Q,R and S be points on the unit circle
such that POQ = ,
POR = +β POS = −β
angles , + β and − β are in standard positions. Points Q,R and S are given by
Q(cos , sin )
R (cos( + β), sin( + β))
S (cos(−β), sin(−β)
Since ’POR and ’SOQ are congruent.
sin sin - cos cos ) cos(
sin 2sin cos cos 2 - 2 2)cos( 2-
)]sin(- -[sin )]cos(- - [cos ) (sin 1] - ) [cos(
2SQ 2PR
SQ PR
2222
Arc lengths PR and SQ, subtends angles +β, +(−β)
Thus, PR = SQ.
Distance b/w (cos , sin ) and (cos(−β), sin(−β)) is
same as distance b/w (cos(+β), sin( +β)) and (1, 0).
2. cos( − β) = cos cos β + sin sin β
cos( + β) = cos cos β − sin sin β
cos( − β) = cos[ + (−β)]
= cos cos(−β) − sin sin(−β)
cos( − β) = cos cos β + sin sin β.
i) If = β, cos2 + sin2 = 1.
ii) If = 0.β = x, cos(−x) = cos x
3. Prove: sin( + β) = sin cos β + cos sin β
sin(+β) = cos[ 2 − ( + β) ]
= cos[( 2 −)− β) ]
= cos( 2 −)cosβ+sin( 2
−) sinβ
sin( + β) = sin cos β + cos sin β
If +β = 2 , cos2 + sin2 = 1
4. Prove: sin( − β) = sin cos β − cos sin β
sin(−β) = sin[ + (−β)]
= sin cos (−β) + cos sin (−β)
sin( − β) = sin cos β − cos sin β
5. Prove tanαanαt1
tanβtanαβ)tan(α
β
ββ)(
β
ββ)(
tantan
tantantan
cos cos
sin sin
cos cos
cos cos cos cos
sin cos
cos cos
cos sin
cos cos
sin sin cos cos cos cos
sin cos cos sin
)cos(
)sin(tan
1
6. Prove tanαanαt1
tanβtanαβ)tan(α
β
β
β
ββ)(
tantan
tantan
)tan(tan
)tan(tantan
1
1
7. Prove cotB cotA
1- cotAcotB B) cot(A
cotcot
cotcot
cotcot
cotcot
tantan
tantan
tan
(tancot
1
11
1
1 2
2
2
2
β
β
β)
β)β)(
8. Expand C) B sin(A
Winglish Coaching Centre, Puduvayal 57 XI Mathematics Made Easy
sinC sinAsinB - sinC cosAcosB
cosC cosAsinB cosC sinAcosB
C) cosAsin(B C) sinAcos(B
C)] (B sin[A C) B sin(A
9. Expand : C) B cos(A
.
B, sinAcos sinC
cosA sinC sinB cosC sinAsinB cosC cosAcosB
sinC sinAcosB - cosC AsinBsin
sinC cosAsinB cosC cosAcosB
C) B cos(A
C B A if,
sinC sinAcosB - cosC AsinBsin
sinC cosAsinB cosC cosAcosB
C) sinAsin(B C) cosAcos(B
)](cos[C) B cos(A
0
02
CBA
10. Expand : C) B tan(A
ACC-BB-A-
CBAC-BA
A
CBA
CBA
tantantantantantan
tantantantantantantanC tanB - 1
tanC tanBtanA - 1
tanC tanB - 1
tanC tanBtan
C) tanAtan(B - 1
)tan(tan
C)] (B tan[A )tan(
1
i) If A+B+C = 0 or π,
C.tan tanAtanB tanC tanB tanAHence
0 C) B tan(A
ii) in case of oblique triangles
tan(x − y) + tan(y − z) + tan(z − x)
= tan(x − y) tan(y − z) tan(z − x
iii) If A + B + C = π/2 tanAtanB + tanB tanC + tanC tanA = 1
11. Evaluate : cos(A + x) − cos(A − x)
x
xx
xxxx
sinAsin 2-
sin sinA cos A cos-
sin sinA cos A cos ) - cos(A - ) cos(A
12. Prove that B sin - A sin B) - sin(A B) sin(A 22
Bsin - Asin
Bsin A)sin - (1 - B)sin - (1 Asin
Bsin Acos - Bcos Asin
sinB) cosA - cosB (sinA sinB) cosA cosB A(sin
22
2222
2222
13. Prove that
Asin-Bcos Bsin - Acos B) - cos(A B)cos(A 2222
Bsin - Acos
Bsin A)cos - (1 - B)sin - (1 Acos
Bsin Asin - Bcos Acos
sinB) sinA cosB (cosA sinB) sinA - cosB (cosA
22
2222
2222
14. Prove that sin2Asin2B B) - (Asin - B) (Asin 22
that cos α + cos β + cos γ = sinα + sinβ + sinγ = 0
2cos(α − β) + 2cos(β − γ) + 2cos(γ− α) +3=0
02
22
2
22
0
22
2222
222
222
2222
22
coscos
coscoscoscoscoscoscos
sinsin
sinsinsinsinsinsinsin
cossincossin
cossinsinsincoscos
sinsincoscossinsincoscos
(cos α + cos β + cos γ)3 + (sinα + sinβ + sinγ)3 = 0
cos α + cos β + cos γ = sinα + sinβ + sinγ = 0
Winglish Coaching Centre, Puduvayal 61 XI Mathematics Made Easy
37. If θ + φ = α and tan θ = k tan φ, then prove that
sin )sin1
1
k
k - (
(k
k -
k
k -
k
k
k
k
)sin
sinsin
)sin(sin
sin
)sin(
)sin(
)sin(
sincoscossin
sincoscossin
tantan
tantan
tan
tantan
tan
tantan
tan
tan
tantan
sin)sin(
1
1
1
1
1
1
38. Show that
cos2A+cos2 B−2cosAcosB cos(A+B) = sin2(A+B)
)cos(coscoscoscos
)sinsincoscos(
coscoscoscos
sincoscossin
coscoscoscos
sincoscossin
coscoscoscoscoscos
sincoscossin
)cos(coscos)cos(
sincoscossin
sincoscossin
)sincoscos(sin
B)]sin(A[
B)(Asin2
2
BABABA
BABA
BABA
BABA
BABA
BABA
BAABAB
BABA
BABA
BABA
BABA
BABA
2
2
2
2
2
2
11
2
22
22
2222
222222
2222
2222
2
3.5. MULTIPLE -SUBMULTIPLE ANGLES
I. Double-Angle Identities
.
A
A A
A
A A -
AA
A A - A
A
A
AA AA
AA A
A A -
A A
A) (A A
.
A A) - - (
A - A -
AAA - A
A) (AA
.
AA A A
A) (AA
Atan 1
A tan -1 cosA
Atan -1
tanA 2sin2A
Atan -1
tanA 2tan2A
Asin 2 -1 2A cos
1 -A 2cos 2A cos
Asin -Acos cos2A
A 2sinAcos 2Asin
2
2
2
2
2
2
22
2
22
2
22
22
22
2
2
22
22
22
2
2
22
1
2
1
1
2
2
cos
sincos
cos
sincos
cossin
sincoscosv)
cos
cos
cossincossin
cossinsiniv)
tantan
tantan
tantaniii)
sinsin
)cos(cos
sinsincoscos
coscosii)
cossincossin
sinsini)
II. Power reducing or Reduction identities
A
AA
A
A
A
A
AA
A A. -
AA
A A -
21
212
212
212
21
221
2
21
212
2
2
2
2
2
2
2
cos
costan
cos
cos
cos
sin
cossin
cossin
coscos
coscos
III. Triple-Angle Identities
Winglish Coaching Centre, Puduvayal 62 XI Mathematics Made Easy
A
AA
AA
A
AA
AAA
AA
AAA
A. A -
A-AA -A -
AAA A - A -
AA A - A
A) A ( A
AA -
AA - AA
AA - A A
AAAAA
AAAA
A) A ( A
2
3
2
2
3
23
2
3
22
22
2
31
3
1
21
1
221
2
23
34
122
212
22
23
43
212
22
21
22
23
tan
tantan
tantan
tan
tantan
tantantan
tantan
)tan(tan iii)
coscos
)cos(coscoscos
sincossincos)cos(
sinsincoscos
coscos ii)
sinsin
sinsin)sin(sin
sinsincossin
sin)sin(coscossin
sincoscossin
sinsin i)
IV. Half-Angle Identities
.A
A -
A
A -
A
A
A -
A
A
A - A
- A
A
A
- A
A
.AA
A
.AA
A
21
21
2
21
22
21
22
2
221
12
2
22
222
222
2
2
2
2
2
2
2
2
22
tan
tancosv)
tan
tansiniv)
tan
tantaniii)
sincos
coscos
sincoscosii)
cossinsin
cossinsini)
V. Prove that AAA-AA 33 444 sincoscossinsin
A.
A A)(
AA)A(
A AA
AA - AARHS
AA A - A A
4
222
222
24
4
44422
33
sin
cossin
coscossin
coscossin
)sin(coscossin
sincoscossinsin
1. Prove that
1010
10
2222
xxxx cos...coscossinsin
1010
10
22
22
2222
22222
222
222
xxxx
xxxx
xxx
xxx
cos...coscossinsin
repeatedly angle half applyAgain
coscossinsinThus,
cos2sinsin
angle, half applyAgain
cossinsin
2. θθθ
θθtan
coscos
sinsin
21
2
θ
θθ
θθθθ
θθθ
tan
)cos(cos
)cos(sincoscos
cossinsin
21
21121
22
3. xx
xxx
cossin
cossinsin
33
22
11
x
xx
xx xx
xxxx xx
xx
xxRHS
22
11
2
21
1
22
33
sin
cossin
cossincossin
)cossincos(sincossin
cossin
cossin
4. Find x such that −π ≤ x ≤ π and cos 2x = sinx
6
5
62
6
5
62
2
1
4
31
012
21
2
2
2
,,Thus
,,1
is sin if
1,- is sin if
2
1or
sin
:formula quadraticUsing
sinsin
sinsin
sincos
x
xx
xx
x
xx
xx
xx
5. Find the values of
i) sin 18◦ ii) cos 18◦ iii) sin 72◦iv) cos 36◦
v) sin 54◦
or Verify that
sin 18◦ = cos 72◦, cos 18◦
= sin72◦ and cos 36◦
= sin54◦
Winglish Coaching Centre, Puduvayal 63 XI Mathematics Made Easy
5
54sin
36cos )36 - sin(90 54sin
72sin
cos18
)18 - sin(90 72sin
5 cos
5
sin cos
sin cosUsing
5
sin - 1 cos
5 sin or
)(
4(4)(-1)- 4 2- sin
sinsin
sinsin
cossin
coscoscossin
cossin
sinsin
2
4
1
4
5210
4
136
4
11
182136
212
4
5210
4
11
1818
4
118
42
0124
3142
342
342
32
3902
3902
9023
905
18
2
2
2
2
2
2
2
3
o
oooo
o
o
ooo
o
oo
oo
o
o
o
o
o
o
AA
-
) - - (
-
. -
) - (
-
6. ooec 20203 seccos
4
202
60604
20202
206020604
20202
202
120
2
3
4
2020
20203
20
1
20
13
0
)sin(
)sin(
cossin
sincoscossin
cossin
sincos
cossin
sincos
cossin
o
oo
oo
ooo
oo
oo
oo
oo
oo
7.
cos
coscos,tantan
a
a
a
a
121
1
2
cos
cos
)cos(sin)sin(cos
)cos(sin)sin(cos
sin)(cos)(
sin)(cos)(
cos
sin
cos
sin
tan
tan
tan
tancos
,tantan
a
a
a
a
aa
aa
a
a
a
a
a
a
a
a
a
a
1
11
11
1
11
1
11
1
11
1
11
1
1
21
1
2
22
22
22
22
22
22
22
22
22
22
22
22
222
22
22
2
22
22
222
2
8. A
A AAAAA
n
nn-
sin
sincos...coscoscoscos
2
2222 1232
A
A
AAAAA
AAAAAA
AAAAAAA
n
n
n-
n-
n-
sin
sin
get we process, the Continuing
cos...coscossinsin
cos...coscoscossinsin
cos...coscoscoscossinsin
2
2
2242
1
22222
1
22222
1
1232
2
1232
1232
9. Find the value ofo
2
122sin
2
22
2
122
2
1
2
1
2
2
1
2
21
21
22
o
sin
45cos45sin
45 take,cos
sin
cossin
oo
o
10. Find sin 2 θ if 13
12sin
169
120
13
5
13
122
2213
51 169
144
cossinsin
cos
Winglish Coaching Centre, Puduvayal 64 XI Mathematics Made Easy
11. If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2
2 tanB) (1 A)tan (1
2 tanB tanA tanB tanA 1
tanB tanA - 2 tanB tanA 1 .
sidesboth on 1 Add
tanB . tanA - 1 Btan tanA .
1 tanB . tanA - 1
tanB tanA
tan45 B) (Atan
BA o45
12. )())()(( oooo 441312111 tan...tantantan is a
multiple of 4 – Prove
2 tanB) (1 A)tan (1
2 tanB tanA tanB tanA 1
tan45 B) (Atan
Let
BA o45
11
22
4
2
22222
231221
4312144111
termsupto....
]tantan
]...tantan][tantan
))((
))(())((oo
oooo
13. Prove )(
-
22
2
1
144
tan
tantansin
)(
-
-
22
2
2
2
2
1
14
1
1
1
22
24
tan
tantan
tan
tan
tan
tan
cossinsin
14. Prove that
2244
tantantan
221
21
2121
1
11
1
1
1
1
11
44
2
2
22
2
22
4
4
4
4
tantan
tantan
tantantantan
tan
)tan()tan(
tan
tan
tan
tan
tantan
tantan
tantan
tantan
tantan
15. Find cos 2A, A lies in the first quadrant, when
63
16 tanA iii)
5
4 sinA ii)
17
15 i)cosA
A
AA
2
2
1
12
tan
tancos
63
16 tanA iii)
4225
3713
25
7
212
289
161
122 22
AAAA sincos
5
4 sinA ii)
coscos
17
15 cosAi)
16. Prove that 64322
17
o
cot
64322
17
13
13
13
1322
13
1322
1
15
151
2
21
1527
2213
2213
21
o
o
o
oo
cot
sin
cos
sin
coscot
17. Prove if
3
3 1
2
13
1
2
1
aa
aa cos,cos
31
2
1
341
2
1
2381
138
1
21
21
3
3
33
3
3
3
3
2
3
3
2
33
cos
coscos
coscos
cos
cos
cos
aa
aa
aa
aa
aa
aa
aa
18. θθθ)(θ)θ)(( nn cottansec...secsec 2214121
θθ
θ
θ
θ
θθθθ
θθθθ
θ
θ
θ
θ
θ
θ
n
n
n
n
nn
n
n
n
n
n
n
n
n
n
n
2
2
2
2
1222
2
2222222
2
2222222
2
2222222
2222
222222
2
21
2
21
2
21
11
22
22
22
32
122222
2
2
tancot
cos
sincot
cos)cossin(
sin
cos
this like Proceedingcos
cos)..cos)(cossin(
sin
cos
cos
cos)..cos)(cos)(sincos(
sin
cos
cos
cos)..cos)(cos)(cos(cos
cos...coscoscos
coscoscoscos
cos
cos...
cos
cos
cos
cos
Winglish Coaching Centre, Puduvayal 65 XI Mathematics Made Easy
19. Prove 3612244848
332
coscoscoscossin)(
36
32
66232
61212234
6122424238
6122448482316
sin
cossin)
coscossin)(
coscoscossin)(
coscoscoscossin)(
20. Prove that θθθ - θ coscoscoscos 520165 35
θθθ - coscoscos
coscoscoscos
coscoscoscos
)cos)cos(cos)cos((
coscoscoscos
)cossincossin(
coscoscoscos
cossin)sinsin(
)cos)(coscos(
sinsincoscos
)coscos(
52016
8866
3468
1816
3468
86
3468
243
1234
2323
23
35
33
335
42
335
42
335
3
23
21. If θ is an acute angle, then find
i)25
1
24
sin,sin ii)
9
8
24
sin,cos
5
32
24
25
11
2
1
12
1
24
2221
2
1
222
222
1
24
222
1
24
422424
2
222
sin
sinsin
sincos
sincossincossin
sidesboth squaring
sincossin
cossincossinsin
23
1
24
222
1
24
cos
sincoscos
above as proceeding
PRODUCT TO SUM AND SUM TO PRODUCT
2
DCsin
2
DC2sin - cosD - cos
2
DCcos
2
DCcos 2 cosD cos
2
D-Csin
2
DCcos 2 sinD - sin
2
D-Ccos
2
DCsin 2 Dsin sin
IVIII,II,I,in eSubstistut2
DCB ,gives(2)(1)
2
DCA ,gives(2)(1)
(2).... D B - A
.....(1), C B A Let
......
sinB sinA cosB cosA B) - cos(A
sinB sinA - cosB cosA B) cos(A
.
sinB cosA - cosB sinA B) - sin(A
sinB cosA cosB sinA B) sin(A
C
C
C
C
I
IVsinB 2sinA - B) -cos(A -B) cos(A
IIcosB cosA 2 B) -cos(A B) cos(A
..IIsinB cosA 2 B) -sin(A -B) sin(A
..IcosB sinA 2 B)-(A sin B) sin(A
1. Express as a sum or difference
.sin 4 5viii)sin 2 cos 5 vii)cos
cos2 sin10 vi)cos2x sin4x v)28 cos iv)sin35
2cos
2sin iii) sin55cos110 ii) cos30sin40 i)
oo
oooo
θθθθ
θθ
xx 3
]cos3 7 [cos2 cos 5 vii)cos
]sin55[sin165
)55 sin(110 - )55 sin(110 sin552cos110 ii)
]sin8 12[sin cos2 sin10 vi)
sin2x] 6x[sin cossinv)
]sin7[sin6328 cosiv)sin35
]sinsin[2
1
22sin
22sin
2cos
2sin iii)
]sin10[sin70
)]30sin(40)30[sin(40cos30sin40i)
oo
oooooo
oooo
oo
oooooo
2
1
2
1
2
12
124
2
1
2
33
2
13
2
12
1
θθ
θθ
xx
xx -
xxxxxx
B) -cos(A B) cos(A cosAcosB 2
B) -sin(A -B) sin(A cosAsinB 2
B) -sin(A B) sin(A cosBsinA 2
Winglish Coaching Centre, Puduvayal 66 XI Mathematics Made Easy
]cos9 [cos.sin 4 5viii)sin
2
1θθ
B) cos(A B) cos(A sinAsinB 2
2. Express as Product
ooooo
ooooo
oo
4cos - 64 cos 34n viii)si75 cos - 35 cos vii)
sin4050vi)sin cos15v)cos6535sin -iv)sin75
2cos -
2 cos iii) cos2 6 cos ii) sin20 sin50 i)
xx
θθ93
20sin 2sin55 75 cos - 35 cos vii)
sinsin
sinsin2
cos -2
cos iii)
C
coscoscos15cos65v)
coscos cos2 6 cos ii)
20sin 2cos5535sin -sin75iv)
5 cos 2sin45sin4050sin vi)
cossin
2
2050cos
2
20 502sin sin20 sin50 i)
oooo
o
oooo
oooo
oooooo
23
23
29
23
29
32
222
93
25402
242
15352
x
xxxx
oo
oo
x
xx
θθ
2
CDsin
2
D2sin cosC -cosD
2
DCcos
2
DCcos 2 cosD Ccos
2
D -Csin
2
D C2cos sinD sinC
2
D -Ccos
2
D C2sin sinD sinC
0
2
ooo
oooooo
30sin 34sin 34sin
] 4cos - 64 cos [ 34sin 4cos - 64 cos 34sin
Prove that
3. 3A cos4
1 A) cosAcos(60 A) - cos(60 oo
A3cos4
13cosA -A cos
4
1
A cosA cos
A cosA cos
A) cos-(1 -A cos
A) sin-60A·(cos cos
A)]A)cos(60-cos(60 cosA[
3
343
243
241
22
oo
4. 3Asin 4
1 A) sinAsin(60 A) - sin(60 oo
3Asin 4
13Asin
sinA 2AsinA cos
sinA ]120 cos - 2A [cos
sinA A) sin(60. A) - sin(60
o
o
2
1
2
12
1
2
12
1
5. tan3A A) tanAtan(60 A) -tan(60 oo
tan3A
A)]A)cos(60-cos(60 cosA[
sinA A) sin(60 A) - sin(60
cos
sin
oo
oo
4343
A
A
6. 16
1 144 cos 108 cos 72 cos 36 cos oooo
16
1
4
15
4
1522
o2o2
oooo
ooooooo
oooo
18sin36 cos
36 cos 18sin 18sin 36 cos
36 - )cos(18018)cos(9018 -cos(90cos36
144 cos 108 cos 72 cos 36 cos
7. 8
1 54sin sin 48 12sin ooo
8
17
721
4
1
7
721
4
1
8
363621
4
1
8
368821
4
1
36821
2
1
2
1
2
12
12
1
41
41
]22sin
sin[
])2-2cos[90
sin[
]2cos1
cossin[
]cos1
cos1cos1sin[
]cos1sin[
)]18sin 54(sin -[1
)18sin 90(sin 54sin -
]54sin 36 c54sin 60 c[
54sin ]60 c36 [cos
o
o
o
o
o
oo
o
ooo
oo
oo
ooo
oooo
ooo
osos
os
8. 16
3 70 cos 50 cos 30 cos 10 cos oooo
16
3
308
3
1034
1
2
3
0
cos
)(cos
]10cos(60 )10-(60 cos 10 [cos 30 cos
]70 cos 50 cos 10 [cos 30 cosoooooo
oooo
o
9. 3
1
oo
oo
15 cos 75 cos
15sin - 75sin
Winglish Coaching Centre, Puduvayal 67 XI Mathematics Made Easy
3
130
22
22
o
oo
oo
oooo
oooo
tan30 cos 45cos
30sin 45cos
1575 cos
1575 2cos
1575sin
15752sin
10. 128
1
15
7
15
6
15
5
15
4
15
3
15
2
15
coscoscoscoscoscoscos
128
1
15
3
15
15
3
15128
1
15
3
15
15
3
15
128
1
15
3
15
15
12
15
16
128
1
15
322
15
322
1542
1542
2
1
15
6
15
3
15
8
15
4
15
2
152
1
15
6
15
3
2
1
15
8
15
4
15
2
15
sinsin
sinsin
sinsin
sinsin
sinsin
sinsin
sin
sin
sin
sin
coscos)coscoscoscos
coscos)cos(coscoscos
11. tan2x.sinsincoscos
cossincossin
x x x - x
x x x - x
432
368
tan2x.cos
sincoscos
sinsincoscos
coscos
]coscoscos[cos
]cossincos[sin
x
xxx
xx x x
x x
x x x x
x x x - x
2
22102
210273
37
73
3979
2121
12. 1. )6 cos - 4(cos )sin - 5(sin
)sin2 (sin8 )3 cos - (cos
θθθθ
θθθθ
1.
)sin5 (2sin )sin2 3 (2cos
)cos3 (2sin5 )sin 2(2sin
θθθθ
θθθθ
13. ).cos (1 sinsinsinsin xxx x x 2232
).cos (1 sin
sincos.sin
sin)sin(sin
xx
xxx
x x x
22
222
23
14. xx x
x x 3
24
24tan
coscos
sinsin
x
xx
xx
3
32
32
tan
coscos
cossin
15. x.xx x x x 3246421 coscoscoscoscoscos
x.xx
xxx
xxx
x x x
324
52
522
64212
coscoscos
)cos(coscos
coscoscos
)cos(cos)cos(
16. sinsinsinsinsinsin 22
11
2
3
2
7
2
θθθθ
sinsin
coscos
coscoscoscos
2
732
1
7432
1
17. 4
1245453030 AA)(-A)(A)(-A)( oooo coscoscoscoscos
4
12
214
1
22
1
4
3
4530
2
2
2222
A
A
A
AA- oo
cos
sin
sin
sincossincos
18. .tancoscoscoscos
sinsinsinsinx
xxx x
xxx x4
753
753
.tancoscos
cossincoscos
sinsin
)cos(coscos
sinsincoscoscoscoscos
cossincossin
)cos(cos)cos(cos
)sin(sin)sin(sincoscoscoscos
sinsinsinsin
x xx
xxxx
xx
xxx
x)x x(xxx x
xxxx
xxxx
xxxxxxx x
xxx x
4242
242124
124
12422
124222122242
2122242
573
573753
753
19. B). tan(A2A) - (4B cos 2B) - (4A cos
2A) - sin(4B 2B) - (4Asin
B). tan(A)cos(
)sin(
cos 2cos
os 2sin
2
2A) 4B2B - (4A
2
2A) - 4B2B - (4A
2
2A) 4B2B - (4A
2
2A) - 4B2B - (4A
BA
BA
c
20. 2sin2A 1
2A cos 4 )15 - (Atan - )15 (A cot oo
2sin2A 1
2A cos 4
sin
cos
]sin[sin
)15 A15 (A c
)15 (Asin )15 - (A cos
)15 (Asin )15 - (Asin - )15 - (A )cos15 (A cos
oo
oo
oooo
21
21
2
22
302
A
A
A
os
Winglish Coaching Centre, Puduvayal 68 XI Mathematics Made Easy
CONDITIONAL TRIGONOMETRIC IDENTITIES
that prove 180 C B AIf o
1. sinC 4sinAsinB sin2C sin2B 2Asin
sinC sinB sinA 4
sinB} sinA {2 2sinC
B)} cos(A - B) - {cos(A sinC 2
B)} A - cos(180 B) - {cos(A sinC 2
cosC} B) - {cos(A 2sinC
cosC 2sinC B) - cos(A 2sinC
sin2C B) - cos(A C) - 2sin(
sin2C B) - cos(A B) 2sin(A
sin2C sin2B) (sin2A
2. 2
Csin
2
Bcos
2
A 4cos 1- cosC - cosB cosA
222
2B A
2B A
2
2B A
22B A
2
22B A
2
22
2B A
2
22
B A22
22
2B A
2B A
sin cos 4cos 1-
1 ] cos [cos 2sin
1 }] - {sin [cos 2sin
1]sin [cos sin 2
1 2sin cos sin 2
1C2sin cos ) -( cos 2
12sin cos cos 2
CBA
C
C
CC
CC
C
C
3. 2
Csin
2
Bsin
2
A 4sin 1- cosC cosB cosA
222
2B A
2B A
2
2B A
22B A
2
22B A
2
22
2B A
2
22
B A22
22
2B A
2B A
sin sin 4sin 1
1 ] cos [cos 2sin
1 }] - {sin - [cos 2sin
1]sin - [cos sin 2
1 2sin cos sin 2
1C2sin cos ) -( cos 2
12sin cos cos 2
CBA
C
C
CC
CC
C
C
4. 8
1
2
Csin
2
Bsin
2
Asin
08
04
02
2
1
u
acb
u
u
uCBA
2B A2
2B A
2B A
2B A2
2B A
2B A
2B A
2
222
cos
cos cos cos
]cos cos [cos-
sin sinsin
81
81
2B A2 cosu
5. 2
31 cosC cosB cosA
23 cosC cosB cosA 1
81 4 1 cosC cosB cosA
1 cosC cosB cosA
6. 4
C-sin
4
B-sin
4
A-sin
2
Csin
2
Bsin
2
Asin
1
444
42442444
42444
44444
442
444
442
442
222222
1
22221
21
21
221
212
CBA
BAABBAABC
BAABC
CABC
CABC
CABBA
CBA
sinsinsin
sinsin)sin(
)]cos())[cos(sin(
)]sin())[cos(sin(
)(sin)cos()sin(
)(sin)cos()cos(
)cos()cos()cos(
7. cosC 2cosAcosB 2 C sin B sin A sin 222
cosC 2cosAcosB 2
1]- B)] (A cos B) -(A [cos C [-2cos
1]- B)}] (A - { cos -B) - (A [cos C [-2cos
1] - C cos -B) - (A [cos C cos 2 - [
1]- 2cos2C B) -(A cos C cos 2- [
1] - C2cos B) -(A cos C) -( cos 2 [
1] -C2cos B)- (A cos B) (A cos 2 [
] Ccos cos2B 2A [cos
] Ccos1 cos2B 1 2A 2cos[1
2
2
21
23
21
23
21
23
21
23
21
23
21
23
21
23 2
22
1
8. cosC 2sinAsinB Csin - B sin A sin 222
C cos Bsin Asin 2
C)] -(Bsin -C) A[sin(Bsin
C)] -(Bsin -A[sinAsin
C)- (Bsin Asin A sin
C)-C)sin(B B sin( A sin
Csin-Bsin A sin
2
2
222
9. 12
Atan
2
Ctan
2
Ctan
2
Btan
2
Btan
2
Atan
1
01
01
1
1
22
22
2B
2C
2C
2B
2B
2A
2B
2C
2C
2B
2B
2A
2C
2B
2A
2C
2B
2A
2B
2C
2C
2B
2B
2A
2B
2C
2C
2B
2B
2A
2C
2B
2A
2C
2B
2A
tantantantantantan
tantantantantantan
tantantan-tantantan
tantantantantantan
tantantantantantan
tantantan-tantantan
tantan CBA
CBA
10. 222 coscos cos 4 sinC sinB sinA CBA
222
222
22B A
2
222B A
2
22B A
2
222B A
2
222B A
22
222B A
2B A
coscos cos 4
}cos cos {2 cos2
}cos {cos cos 2
)} - sin( {cos cos 2
}in {cos cos2
cossin cos cos2
cossin cos )-2sin(
cos2sin cos 2sin
CBA
BAC
BAC
BAC
CC
CCC
CCC
CC
s
2
2
Winglish Coaching Centre, Puduvayal 69 XI Mathematics Made Easy
11. C. cos cosAcosB 2-1 C cos B cosAcos 222
cosC 2cosAcosB 1
1]- B)] (A cos B) -(A [cos C [-2cos
1]- B)}] (A - { cos -B) - (A [cos C [-2cos
1] - C cos -B) - (A [cos C cos 2 - [
1]- 2cos2C B) -(A cos C cos 2- [
1] - C2cos B) -(A cos C) -( cos 2 [
1] -C2cos B)- (A cos B) (A cos 2 [
] Ccos cos2B 2A [cos
] Ccos1 cos2B 1 2A 2cos[1
2
2
21
23
21
23
21
23
21
23
21
23
21
23
21
23 2
22
1
12. sinC 4sinAsinB C) - B sin(A B) - A sin(C A) - C sin(B
Csin Bsin Asin 4
2Csin 2Bsin 2Asin
2C) -(sin 2B)-sin( 2A)-sin(
C) -C-(sin B)-B-sin( A)-A-sin(
C) - B sin(A B) - A sin(C A) - C sin(B
13. If A + B + C = 2s, then prove that sin(s − A) sin(s − B) + sins sin(s − C) = sinAsinB.
sinAsinB
]2
cos2
A-B[cos2
1
]2
cos2Ccos
2cos
2A-B[cos
2
1
]2
C-scos2Ccos
2B)(A-2s
cos2
A-B[cos2
1
]2
C-sscos2
Cs-scos
2BsA-scos
2Bs-A-s[cos
2
1
C) - sin(s sin B) - A)sin(s-sin(s
AB
ABC
s
2
14. If x + y + z = xyz, then prove that
. z2 - 1
2z
y2- 1
2y
x2 - 1
2x
z2 - 1
2z
y2- 1
2y
x2 - 1
2x
. - z
z
- y
y
- x
x
- z
z
- y
y
- x
x
CBACBA
nC BA
nCB A
CB A C BA
xyzzyx
CzB,yA,x
222222 1
2
1
2
1
2
1
2
1
2
1
2
222222
2222
tantantantantantan
tantantantantantan
Now
tantantanLet
If A + B + C =2
, prove the following
15. cosC 4cosAcosB sin2C sin2B 2Asin
sinC} B) - {cos(A C2cos
cosC 2sinC B) - cos(A 2cosC
sin2C B) - cos(A C) - 2sin(
sin2C B) - cos(A B) 2sin(A
sin2C sin2B) (sin2A
2
cosC 4cosAcosB
cosB} cosA {2 2cosC
B)} cos(A B) - {cos(A C cos 2
B)} A (- sin( B) - {cos(A cosC 2 2
16. sinC 4sinAsinB 1 cos2C cos2B 2A cos
Csin 4sinAsinB 1
1 B)] (A cos B) -(A [cos C2sin
1B)} (A - {sin -B) - (A [cos C2sin
1C]sin -B) - (A [cos Csin 2
1 C2sin B) -(A cos Csin 2
1C2sin B) -(A cos C) -( cos 2
1C2sin B)- (A cos B) (A cos 2
2
2
22
2
If ABC is a right triangle and if 2
A , then
prove that
17. 1 C cos B cos 22
1
0
0
0
90
2222
CC C B
CB
CB
CB
CB
CB
o
o
o
o
cossincoscos
sincos
)cos(cos
18. 1 C sin B sin 22
1
2222
CC C B sincossinsin
above as Proceeding
19. 2
Csin
2
Bcos22 1- cosC - cosB
2sin
2cos
2 4cos 1-
1 ]2
B A cos 2
B A [cos 2
2sin
1 }]2
B A - 2
{sin 2
B A [cos 2
2sin
1]2
sin 2
B A [cos 2
sin 2
1 2
22sin 2
B A cos 2
sin 2
1C22sin 2
B A cos )2
-2
( cos 2
12
22sin 2
B A cos 2
B A cos 2
cosC - cosB CosA cosnsider,
CBA
C
C
CC
CC
C
C
2
Csin
2
Bcos22 1- cosC - cosB
2
Csin
2
Bcos
4 4cos 1- cosC - cosB
2Cos
Winglish Coaching Centre, Puduvayal 70 XI Mathematics Made Easy
TRIGONOMETRIC EQUATIONS
Principal Solution
Smallest numerical value of unknown angle satisfying the equation in the interval [−π, π] is
taken for defining principal solution.
Principal values
Function interval Quadrant
Sine ],[ 22 I or IV
Cosine [0, π] I or II
tangent ],[ 22 I or IV
General solution of trigonometric equations.
1. Solve the equation sin θ = k (−1 ≤ k ≤ 1) :
Let sin = k, numerically smallest angle
Z., n)(-n
Znn
Znn
Znn
Znn
-
n
1
222
02
1122
122
02
022
2
0
,Combining
)...(,
,sinor
)...(,)(
,)(cosEither
sincos
sinsin
sinsin
2. Solve equation of the form cos θ = k (−1 ≤ k ≤ 1)
Z., nn
Znn
Znn
Znn
Znn
2
222
02
122
02
022
2
0
,Combining
)...(,
,sinor
)...(,
,sinEither
sinsin
coscos
coscos
3. Solve equation of the form tan θ = k (−∞ < k < ∞)
Znn
Znn
,
,
)sin(
sincoscossincos
sin
cos
sin
tantan
0
0
4. Solve an equation of the form a cos θ+b sin θ = c
znn
znn -
) -(
ba
c) -(
r
c ) -(
cr r
c baa
bbar
r , b r a
,
,
(say)coscos
cos
cos
sinsincoscos
sincos
tan,
sincos Take
2
2
22
22
5. Find the principal solution of
2
1iii)cos2,- ii)cosec
1sin i)
2
3
3
00
0
6
6
02
coscos
quadrant Iin lies valueprincipal
],in[ lies cos of valueprincipal,2
1iii)cos
6
6sin sin
quadrant IVin lies valueprincipal2
1- sin
2,- sin
1ii)
sinsin
quadrant Iin lies valueprincipal
],in[ lies sin of valueprincipal1
sini)
2
22-
6. Find the general solution of
i)2
- sin 3
ii) sec θ = −2 iii) tan θ =√3
Winglish Coaching Centre, Puduvayal 71 XI Mathematics Made Easy
Znnπ
Znnπ
Znnπ
n
n
n
,3
(-1) or
,3
-(-1)
,(-1)
:solution general
3-sin sin
1
ii) sec θ = −2
Znnπ
Znnπ
,3
, solution general
3cos cos
3-cos cos
2
1-cos
22
2
2
iii) tan θ =√3
Znnπ
Znnπ
,3
,
:solution general
3tan tan
7. Find the principal solution and general solution
33
1
2 coti)itanii)
1sin i) i
42
1sin i)
Znnπ
Znnπ
Znnπ
n
n
n
,3
(-1) or
,3
-(-1)
,(-1)
:solution general
1
66
03
1
tantan
tanii)
Znnπ
Znnπ
,3
, solution general
3coti)ii
Znnπ
,6
solution general
tantan
tan
66
03
1
8. Solve the following equations for which solutions lies in the interval 0◦ ≤ θ < 360◦
i) sin4 x = sin2 x
220
2
2
01
0
01
0
22
2
2
22
24
,,,
sinsin
sinor
sin Either,
)(sinsin
sinsin
x
x
x
x
x
x
xx
x x
ii) 2 cos2 x + 1 = −3 cos x
,,
cos
cosOr
coscos
)(cosEither,
)cos)((cos
cos 3 cos 2 2
3
4
3
23
2
2
1
012
01
0121
01
x
xx
x
xx
x
xx
xx
iii) 2 sin2x+1 = 3sinx
6
5
62
62
1
012
01
0121
01
,,
sin
sinOr
,sinsin
)sin(Either,
)sin)(sin(
sin 3 sin 2
22
2
x
xx
x
xx
x
x
xx
iv) cos 2x = 1− 3 sinx.
,
possible Notsin
sinOr
,
sinEither,
)sin(sin
sin 3 sin 2
sinsin2
02
3
032
0
0
032
0
3121 2
x
x
x
x
x
xx
xx
x- x
Solve the following equations:
1. 3 cos2 θ = sin2 θ
Winglish Coaching Centre, Puduvayal 72 XI Mathematics Made Easy
znn
znn
-
,
,
)cos(cos
cos
cos
cos
coscos
3
32
3
2
22
22
2
2
12
4
1
2
214
1
13
2. sin x + sin5x = sin3x
Z n ,6
n x (or)3
n xsolution general
Z n ,6
n x 3
2n 2x
3cos 2x cos , 2x cos If
. Z n ,3
n x
. n 3xthen 0, 3xsin If
2x cos (or) 0 3xsin Either
0 1) - 2x cos (2 3xsin
sin3x 2x cos sin3x 2
. sin3x sin5x xsin
2
1
2
1
3. cos x + sinx = cos2x + sin2x
znn
x
n
znnxnx
xx - x x -
x xx x
xx
xx
xx
x
xxx
xxxx
xxxxxxxx
,
,tan
cossin
]cos[sin Or
,
sin Either
]cos[sinsin
sincossinsin
sincossinsin
sinsincoscos
sincossincos
63
2
1
0
22
0
0
22
22
22
22
423
23
23
23
23
23
2
23
23
2
223
223
22
22
22
22
4. sin 9θ = sinθ
znn
n
(or)
znnn
-
,
sin
)()(
cosEither
sincos
sinsin
sinsin
44
0410
122
125
05
0452
09
9
5. tan 2x = −cot )( 3x
znnx
znxnx
x x
x x
x -x
,
,
)tan(tan
)tan(tan
)cot(tan
65
65
65
32
3
2
2
2
2
6. sin x − 3 sin2x + sin3x = cos x − 3 cos 2x + cos3x
Z. n ,
tan
cossin
cossincossin
coscossin
coscoscossincos
coscoscossinsinsin
coscoscossinsinsin
82
12
22
032022
03222
23222322
233233
323323
n x
x
x x
x -ce x x -
)x-x)(x-(
x x - xx x - x in
xx -xx x - x
xxx - x x x -
7. 0 1 - 1)tan - 3( tan 3 2
znn
or
znn
4-
-1tanthen 0 1tan If
0 1tan
,
tan
3
1tan 0,1 -tan 3(If
01-tan 3(Either
0 1) 1)(tan -tan 3(
0 1 - tan-tan 3 tan 3
0 1 - 1)tan - 3( tan 32
2
6
6
8. sin x + cos x = 1 + sinx cos x
Znnxnx
nx
x
xx
xx
x x
ttt
t
txx txx
txx
,,
coscos
coscossincos
cossin
cossin
)(
1teqn,given in sub
cossincossin
cossinLet
222
42
4
44
2
1
44
12
1
2
12
1
01012
2
1
2
121
22
2
22
9. √ 3 sinθ − cos θ =√2
Winglish Coaching Centre, Puduvayal 73 XI Mathematics Made Easy
Znn
n
. bar; c b ; -a
46
46
46
2
1
66
2
2
2
1
2
3
2231 22
n
n
(-1)
(-1)
sinsin
sincoscossin
cossin
10. sin θ + cos θ =√2
Znn
n
,)(
coscos
coscossinsin
cossin
418
024
04
144
2
2
2
1
2
1
11. sin θ +√3 cos θ = 1
Znn
n
,
coscos
coscossinsin
cossin
632
32
6
36
2
1
66
2
1
2
3
2
1
12. cot θ + cosecθ = √3
Znn
n
,
coscos
coscossinsin
sincos
sincos
sinsin
cos
33
22
3
22
3
33
2
1
33
2
1
2
3
2
1
13
31
13. cos 2θ = 4
15
10
2022
362
n
n
o ]5:Problem62,:pagesee[coscos
14. 2 cos2 x − 7 cos x+3 = 0
32
2
1
012
3
0123
0372 2
nxx
x
x
xx
xx-
cos
cos Or,
possibleNotcosEither,
)cos)((cos
coscos
15. 2 cos2 θ + 3sinθ − 3 = 0
61
2
1
012
1
1
01
0121
01
03
03
n
n
n
n
)(sin
sinOr
)(,sinsin
sin
)sin(Either,
)sin)(sin(
sin 3 sin
sin 3 sin-2
sin 3 cos 2
22
2
2
2
16. 2 sin2 x + sin2 2x = 2
Z , n nx
x
x or
Z, n)n (x
x
x - x
x)x (x
xx
4
4
2
12
12
0
01
222
222
22
2
22
22
22
sinsin
sin
cosEither
][sincos
cossinsin
sinsin
17. sin 5x − sin x = cos3x
Z n,12
n or6
1)(2n solution
Z n ,12
n x 6
2n 2x
6sin 2xsin
2xsin (or)
Z n ,6
1)(2n x ,2
1)(2n 3x
0 3x cosEither
0 1) - 2x(2sin 3x cos
cos3x 3x cos sin2x 2
. 3xcos sinx 5xsin
xx
2
1
18. cos θ + cos3θ = 2cos2θ
0122
02222
0223
) -(
-
-
coscos
coscoscos
cos]cos[cos
Winglish Coaching Centre, Puduvayal 74 XI Mathematics Made Easy
Znn
-or,
Z, n)n (
, n Z )n (
,
coscos
cos
2
1014
12
2122
02
19. sin θ + sin3θ + sin5θ = 0
Z n ,3
n (or)3
n xsolution general
Z n ,3
n 3
2n 2
3cos 2 cos , 2 cos If
. Z n ,3
n
. n 3then 0, 3sin If
2 cos (or) 0 3sin Either
0 1) 2 cos (2 3sin
sin3 2 cos sin3 2
sin3- sin5 sin
2
2
2
1
21
20. sin 2θ − cos 2θ − sin θ + cos θ = 0
Znn
n
Znnn
632
4
1
0
22
0
0
02
0
23
tan
sincos or
,
sin Either
)sin(cossin2
sinsinsin2cos
)cos cos2()sin 2(sin
23
23
23
2
23
23
2
223
223
21. tan θ + tan )( 3 + tan )( 3
2 =√3
183
6
2
3
2
3
2
3232
3
3
32
3
3
63
3
1
31
3
331
83
331
8
331
3
31
3
311
3
n
n
tan
tan
tantan
tan
tantantan
tan
tantan
tan
tan
tan
tantan
tantan
tantan
tantan
tantantan
)(tan)(tantan
PROPERTIES OF TRIANGLE
1. Law of Sines
In any triangle, lengths of the sides are proportional to the sines of the opposite angles
RC
c
B
b
A
a2
sinsinsin
RC
c
B
bA
a R
R
a A
R
a - A) (or
BD
BCBDC
- A BDC
BAC BDC
D. circle ato meet theoduce BO t
.
R A
a
R
BC
A
a
R A
a
R
aA or
BD
BCBDC
BCD
ABDC
BAC BDC
.
o
o
o
o
o
2
2
2
2180
180
180
2
1
290
2
2
90
sinsin simillarly
sin
sin
sin
sin
Pr
sin
sinsin
ABC. of BC side theon is O
.sin
sin
sin
]semicircle ain .[angle
segment] samein [angles
D at circle the meet to BO Produce
.
obtuse is A :III Case
angle right is A :II Case
acute is A :I Case
2. Napier’s Formula
22
2222B
ac
acAC
A
cb
cbCBC
ba
baBA
cottan
cottan,cottan
Winglish Coaching Centre, Puduvayal 75 XI Mathematics Made Easy
22
222
2222
222
22
2
2
222
22
2
22
22
22
22
BAC
ba
ba
CBAC
CBAC
CBABA
C
C
BA
BA
C
BRAR
BRARC
ba
ba
BRbARa
RB
bR
A
a
BABA
BABA
tancot
cottantan
cottancot
cottancot
cotcossin
sincos
cotsinsin
sinsin
cotsinsin
sinsincot
sinsinsinsin
Law sine
3. Law of Cosines
ab
-cabC
ca
-bacB
bc
-acbA
2
22222
222222
cos
cos,cos
bc
-acbA
Cab - b a c
Cab - b a
Cab - CCb a
CabC - baC b
C)(a - b C) (b c
C a - b
BC - DC BD
C b AD
C AC
AD
BDAD c
BDADABΔABD
2
2
2
2
2
222
222
22
2222
22222
222
222
222
cos
cos
cos
cos)cos(sin
coscossin
cossin
cos
sin
sin
,In
ABCin BCAD Draw
4. Projection Formula
A b B a iii) c
C a A c ii) b
B, c C b i) a
coscos
coscos
coscos
B c C b a
C)b (B)c (
ACAC
DCAB
AB
BD
DC BD
BC a
BCDraw AD
BC ABC,a In
coscos
coscos
5. Area of the Triangle
C.ab
C b AD C AC
AD
Bac
Abc C ab
sin
height base ,Thus
sin sinADC,In
BCAD draw ABC,In
sin
sinsin triangle of Area
21
21
21
21
21
6. Area of the segment of a circle
θθ (θ(θ r
θ rθ - r
sin
sin
OABof Area -Area Sector ABD segment of Area
221
2212
21
7. Half-Angle formula
bc
c) - b)(s - s(s - a)(s
bc
s(s - a)
bc
- c)(s - b)(s
AAA
bc
- c)(s - b)(s
bc
c)s - b)(s - (
bc
bcbaccba
bc
cbacba
bc
cba
bc
acbbc
bc
acb
A
AA
bc
- c)(s - b)(s Aiii)
s(s - a)
- c)(s - b)(s Aii)
bc
s(s - a)Ai)
cba s
s In ABC
2
2
222
4
2222
4
22
4
4
4
2
21
2
1
2
1
22
2
22
2
22
222
222
2
cossinsin
))((
))((
)(
cos
sinsin
sin
tancos
ABCofΔf perimeter-semiif
8. Heron’s formula
c) - b)(s - s(s - a)(s
ab
s(s - c)
ab
- a)(s - b)(s ab
ab
C.ab
CC
2221
21
2 cossin
sin
Winglish Coaching Centre, Puduvayal 76 XI Mathematics Made Easy
In a ABC, prove that 1. b2 sin 2C+c2 sin 2B = 2bc sinA
A.sin 2bc
sinA 2R
c
2R
b 8R
sinA sinC sinB 8R
A) - (sin sinC sinB 8R2
C) sin(B sinC sinB 8R
cosB) sinC cosC (sinB sinC sinB 8R
CsinBcosBsinsinCcosC B2sin 4R
sin2B Csin 4Rsin2C Bsin 4R
sinsinsinsinsinsin
Law, Sine
2
2
2
2
222
2222
2
222
2
CR B; c R A; b R a
R C
c
B
b
A
a
2. 2
Acos
a
c - b
2
C - Bsin
2
C - Bsin
sin
2cos
2
C - Bsin
cos
cossin
2
C Bcos
2
C - B2sin
cos2RsinA
2RsinC - 2RsinB
2
Acos
a
c - b
sinsinsinsinsinsin
:Law Sine
2
2
2
222
2
222
2
A
A
A
AA
A
CR B; c R A; b R a
R C
c
B
b
A
a
3. If the three angles in a triangle are in the ratio 1 : 2 : 3, then prove that the corresponding sides are in the ratio 1 :√3 : 2.
131
22
3
2
1
::::
::
90 sin60 sin:30sin::
90sin60sin30sin
:Law Sine
90 ,60 ,30 angles
30
180. 32
3 ,2 , angles the Let
ooo
ooo
ooo
o
cba
cba
cba
4. (b+c) cosA+(c+ a) cosB + (a + b) cosC = a + b + c = b cosA+c cosA+c cosB+a cosB + a cosC+b cosC = b cosC +c cosB+c cosA+a cosC+b cosA+a cosB = a + b + c [by projection formula]
5. The angles of a triangle ABC, are in Arithmetic
Progression and if b : c = √3 :√2, find A.
o
oo
o
o
A
CBA
CC
C
B
c
bC
c
B
b
: b : c
B
B
BCABB
CAB
CBA
75
4560
2
3
23
60
1803
2
2
)(
sin
sin
sin
sinsinsin
LawSin
A.Pin are,,
6. B(A - C)
C(A - B)
ca
ba
coscos
coscos
1
122
22
CR B; c R A; b R a
R C
c
B
b
A
a
sinsinsinsinsinsin
Law Sine
222
2
B(A - C)
C(A - B)
C)R(A)R(
B)R(A)R(
ca
ba
coscos
coscos
C) - cos(A C) cos(A - 1
B) - cos(A B) cos(A - 1
C) sin - A(cos- 1
B) sin - A(cos- 1
Csin Acos - 1
B sin Acos - 1
C sin A sin
B sin Asin
sinsin
sinsin
1
1
22
22
22
22
22
22
22
22
22
22
22
22
7. Cb - a
Bc - a
C
B
cos
cos
sin
sin
C
BCR
BRAc
Ab
C - a AcCa
B - a AbBa
Cb - a
Bc - a
sin
sinsin
sincos
cos
cos)coscos(
cos)coscos(
cos
cos
2
2
8. a cosA + b cosB + c cosC = 2a sinB sinC.
CBa
CBR
CBAR
CB A R
CCRBBRAAR
C cBbA a
Ra
sinsin
)sinsin(
)sinsinsin(
)sinsin(sin
cossincossincossin
coscoscos
2
4
4
222
222
2
Winglish Coaching Centre, Puduvayal 77 XI Mathematics Made Easy
9. C)(A
(A - C)
b
-ca
sin
sin2
22
C)(A
(A - C)
C)(A
(A - C)CA
CA
CACA
B
CACAB
CA
BR
CRAR
b
-ca
CR B; c R A; b R a
R C
c
B
b
A
a
sin
sin
sin
sin)sin(
])[sin(
)sin()sin(
sin
)sin()sin(sin
sinsin
sin
sinsin
sinsinsinsinsinsin
:Law sine
2
2
2
22
22
2222
2
22
4
44
222
2
10. 222222 - ba
(A -B)c
- ac
(C - A)b
- cb
(B - C)a sinsinsin
R- ba
(A -B)c
- ac
(C - A)b
- cb
(B - C)a
R- ba
(A -B)c R
ACAC
(C - A)B
R- ac
(C - A)b R
A
A
R
CBCB
(B - C)A
R
CRB- R
(B - C)A R
- cb
(B - C)a
2
1
2
12
1
2
12
1
2
1
2
144
2
222222
22
22
222222
sinsinsin
sin
)sin()sin(
sinsinsin
)sin(
sin
)sin()sin(
sinsinsinsin
sinsinsin
11.
22
BABA
ba
bacottan
22
222
222
22
22
BABA
BABA
BABA
BA
BABRAR
BRAR
ba
ba
cottan
sincos
cossin
sinsin
sinsinsinsin
sinsin
12. Derive cosine formula using the law of sines
CR B; c R A; b R a
R C
c
B
b
A
a
sinsinsinsinsinsin
:Law Sine
222
2
cosAsinC 2
A) - sin(C A) sin(CsinC sinB 2
A)] - sin(C [sinB sinBsinC sinB 2
A) - sin(C A) sin(C Bsin
(2RsinC) (2RsinB) 2
(2RsinA) - (2RsinC)(2RsinB)
2
222222
bc
- acb
2
13. Derive Projection formula from Law of sines,
BcCba
BCRCBR
CBR
CBR
AR a
R A
a
coscos
cossincossin
]sin[
])[sin(
sinsin
,Law Sine
22
2
2
2
2
14. Derive Projection formula from Law of cosines.
aCbBca
a
a
-cab-bca
ab
-cabb
ac
-bcacCbBc
ab
-cabC
ac
-bcaB
coscos
coscos
cos,coslaw,cosine
2
2
2
22
22
2
222222
222222
222222
15. Using Heron’s formula, show that equilateral triangle has maximum area for fixed perimeter. Let ABC be a triangle with constant perimeter 2s.
c) - b)(s - s(s - a)(s
is maximum, when (s−a)(s−b))(s−c) is maximum
27
s c) - b)(s - a)(s - (s
A.M. G.M
3
273
33 s (s - c) (s - b) (s - a) )(s - c)(s- a)(s-b
Equality occurs when s − a = s − b = s − c.
when a = b = c, maximum of(s − a)(s − b))(s − c)=27
3s
For fixed perimeter 2s, area is maximum if a = b = c
Maximum area 3327
23 sss sq.units.
16. In a ABC, if C) - sin(B
B) - sin(A
sinC
sinA , prove that a2, b2,
c2 are in Arithmetic Progression.
B) B)sin(A - sin(AC) - C)sin(Bsin(B
C) - sin(B
B) - sin(A
B)sin(A
C)sin(B
C) - sin(B
B) - sin(A
B])[A-sin(
C])[B-sin(
Winglish Coaching Centre, Puduvayal 78 XI Mathematics Made Easy
222
2222
2
222
cab
bacb
R
c
R
b
R
a
BACB
sinC,sinB,sinA law, sine
sinsinsinsin 2222
a2, b2, c2 are in Arithmetic Progression.
17. If sinB 2
sinA cosC , show that triangle is isosceles.
isoceles Hence
sinB 2
sinA cosC Now
cos
:lawcosine
sin,sin,sin
: law sine
ac
cb
a-cab
b
R
R
a
ab
-cab
ab
-cabC
R
cC
R
bB
R
aA
22
2222
222
222
2
2
22
2
222
18. Prove that
2
2B - C
a c b cos if A = 60o
22
2602
22
602
2
222
2
22260
B - Ca cb
B - Ca
B - Ca
cb
B - CAa
cb
B - CCBacb
CBaAcb
AbBaAcCacb
o
o
o
cos
coscos
coscos
coscos
)coscos(cos)(
)cos(coscos)(
coscoscoscos
19. 2
2 2 Ac) (b C) Ba( sincoscos
22
211
1
2
2
Ac) (b
Acb
Acb
AcbAbc
CaB aC) Ba(
sin
)sin)((
)cos)((
coscos
coscoscoscos
20. 22
A c) (b B
Aa sinsin
BA
a
A c) (b
2
2
sin
sin
21. (a2 − b2 + c2)tanB = (a2 + b2 − c2) tanC
CabC
CaCabaC) cb (a
Cabc
Cbca
BcaB
BccBcaB )cb(a
cBcaba
sincos
sin)cos(tan
sin
sin
sincos
sin)cos(tan
cos
2
2
2
2
2
2
2
22222
22222
222
22. An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park
to be developed must be of maximum area. Find out the dimensions of the park. For fixed perimeter, and maximum area the triangle should be equilateral
40
120
120
120
a
a
cba
cba
3
since
Perimeter
23. A rope of length 12 m is given. Find the largest
area of the triangle formed by this rope and find the dimensions of the triangle so formed.
sq.m
s Area Max
3
Perimeter
2
34
33
6
33
4
12
12
2
a
a
24. Government plans a circular zoological park of diameter 8 km. Find separate area in the form of a segment formed by a chord of length 4 km is to be
allotted for a veterinary hospital in the park.
234
22
332
2
3
38
3
2
1
8
m
rr
][
segment of Area
2(4)(4)
4- 4 4cos
]sin[
sin2
1
2
1
OAB.of Area - sector of Area segment of Area
m4r, .AOB
park. circular of centre O
chord AB Let
222
Winglish Coaching Centre, Puduvayal 79 XI Mathematics Made Easy
3.11.APPLTICATION OF TRIGNOMETRY
Working Rule: In a right triangle, two sides determine third side
via Pythagorean theorem and one acute angle
determine other using the fact that acute angles in
a right triangle are complementary.
If all sides of triangle are given, use either cosine
formula or half-angle formula to calculate all the
angles of the triangle.
If two angles and one of sides opposite to given
angles given, use sine formula to find other sides.
If two sides and included angle given, use law of
cosines to calculate other side and other angles of
triangle. In this case we have a unique triangle.
Solving an oblique triangle require that length of
atleast one side must be provided.
SOLVED PROBLEMS
1. In a ABC, a = 3, b = 5 and c = 7. Find the values
of cos A, cosB and cos C
.
2
1- cosC ,
14
11 cosB
14
13
2(5)(7)
32 - 72 52
a - c b cosA
:formula Cosine By222
bc2
2. In ABC, A = 30,B = 60,c = 10, Find a and b.
3590
10
60
590
10
30
90
10
6030
bb
aa
baC
c
B
b
A
a
oo
oo
ooo
oooo
sinsin
sinsin
sinsinsin
sinsinsin law sine
90 90 - 180 C ,60 B,30 A o
3. If the sides of a ABC are a = 4, b = 6 and c = 8,
then show that 4 cosB + 3cosC = 2.
243
486
B C
B C
aB c C b
coscos
coscos
coscos
:Formula Projection
4. In a ABC, if a = 2√2, b = 2√3 and C = 75, find
the other side and the angles
o
o
o
BCAB
AA
A
c
ab
- cba
C
60180
452
1
13234
1334
13234
834812
13222
13
22
13
2
75222
)(
cos
)(
)(
)(cos
)(68
c - 12 8
coscos
2
5. In a ABC, if a =√3 − 1, b =√3 + 1 and C = 60,
find the other side and other two angles.
o
o
o
BB
AA
A
c
C
10575180
1522
13
1323
233
1323
332
13232
13613
62
1
1313
1313
60
0
22
22
cos
)(
)(
)(
)(
)()(cos
))((2
c - )( )(
coscos2
6. Find area of triangle : sides 13 cm, 14 cm,15 cm.
sq.cm. 84
15) - 14)(21 - 13)(21 - 21(21
c) - b)(s - a)(s - s(s triangle of Area
cm. 21 15 14 13
c b a s
2
2
7. In a ABC, if a = 12 cm, b = 8 cm and C = 30,
then show that its area is 24 sq.cm.
sq.cm
sin
sin
24
308122
12
1
o
Cab
8. In a ABC, if a = 18 cm, b = 24 cm and c = 30 cm,
then show that its area is 216 sq.cm.
Winglish Coaching Centre, Puduvayal 80 XI Mathematics Made Easy
sq.cm. 216
61836
c) - b)(s - a)(s - s(s triangle of Area
cm. 33 24 18
c b a s
12
62
02
9. InABC prove that abc
CcBbAa28
coscoscos
Refer to problem 8, page : 76
abcCcBbAa
abaca
CB a CcBbAa
28
222
2
coscoscos
sinsincoscoscos
10. In any ABC, prove that the area = A
- acb
cot4
222
A
- acb
AA
- acb
Abc
A
- acbbc
bc
- acbA
cot
sincos2
1
sin2
1Triangle of Area
cos
coslaw, cosine
4
2
2
2
222
222
222
222
11. A boat travels 10 km from the port towards east
and then turns 60 to its left. If the boat travels
further 8 km, how far from the port is the boat?
km612 BP
. 244km
120 cos 8 10 2-8 10 BP
formula, cosineUsing
distance. required BP
o222
12. Two cell phone towers located at 6 km apart,
east to west. A cell phone is north of highway.
The signal is 5 km from the first tower and √31
km from the second tower. Determine the
position of the cell phone north and east of the
first tower and how far it is from the highway.
o
22
60
cos
cos 60 -3625 31
cos 6 5 2 - 6 531
2
1
2
km3 5
5sin60 .
sin60
o
o
highway fromposition sphone’ of distance
2
5
x
x
x
13. Two radar stations located 100km apart, detect a
fighter aircraft between them. Angle of
elevation measured by first station is 30. Angle
of elevation measured by the second station is
45. Find altitude of the aircraft at that instant.
)(
sin
)(
sinsin
105 ) 45 (30 -180 A oooo
1350
30
13100
13
22100
1
2
105
100
45
x
a
x
a
a
a
o
oo
14. Two soldiers A and B in two underground
bunkers spot an intruder atthe top of a hill. The
angle of elevation of the intruder from A and B
to the ground level in the eastern direction are
30 and 45. If A and B stand 5km apart, find the
distance of the intruder from B.
13
25
13
225
1
2
105
5
30
BC
BC
BCoo sinsin
105 ) 45 (30 -180 A
detectionduring intruder ofposition C oooo
15. A researcher wants to determine the width of a
pond from east to west. From a point P, he finds
the distance to eastern-most point of the pond to
be 8 km, distance to the western most point from
P to be 6 km. If the angle between the two lines
of sight is 60, find the width of pond.
km132 a
. 52km
60 cos 8 6 2-6 8 a
A cos c2-c b a
formula, cosineUsing
o222
222
b
16. Two Navy helicopters A and B are flying over
the Bay of Bengal at same altitude to search a
missing boat apart 10 km from each other. If the
Winglish Coaching Centre, Puduvayal 81 XI Mathematics Made Easy
distance of boat from A is 6 km and if the line
segment AB subtends 60 at the boat, find the
distance of the boat from B.
km192 a
. 76km
60 cos 1 6 2-10 6 a
A cos c2-c b a
formula, cosineUsing
o222
222
0
b
17. A surveyor observes two extremities A and B of
the tunnel to be built from point P in front of
mountain. If AP = 3km, BP = 5 km APB = 120,
find the length of the tunnel to be built.
7km a
. 49km
120 cos 8 6 2-5 3 a
A cos c2-c b a
formula, cosineUsing
o222
222
b
18. A farmer purchase a triangular shaped land
with sides 120feet and 60feet and angle included
between these two sides is 60. If the land costs
Rs.500/sq.ft, find amount he needed to purchase
the land. Also find the perimeter of the land.
1558800500318005001
31800
26900
131333333030
30330303303309033090
33090
2
2
Cost,.
))()()((
))(())((
))()((
36060120s
360 a
10800
7200-3600 14400 a
formula, cosineUsing 2
Rsft
csbsass
19. A fighter jet has to hit a small target by flying a
horizontal distance. When the target is sighted,
angle of depression is 30. If after 100 km, target
has an angle of depression of 45, how far is the
target from the fighter jet at that instant?
Refer to Problem 13 , page 79
20. A plane is 1 km from one landmark and 2 km
from another. From the planes point of view the
land between them subtends an angle of 45.
How far apart are the landmarks?
km 22-5 a
. km22-5
45cos 2 1 2-2 1 a
A cos c2-c b a
formula, cosineUsing
o222
222
b
21. A man starts his morning walk at a point A
reaches two points B and C and finally back to A
such that A = 60 and B = 45 , AC = 4km in
the ABC. Find the total distance he covered
during his morning walk.
)(DistanceTotal
)(
sinsin
sinsin
132624
132
22
1324
45
4
75
62
2
324
45
4
60
75105180
AB
AB
BC
BC
C
oo
oo
o
22. Two vehicles leave the same place P along two
roads. One vehicle moves at an average speed of
60km/hr and the other vehicle moves at an
average speed of 80 km/hr. After half an hour
vehicle reach the destinations A and B. If AB
subtends 60 at the initial point P, then find AB.
km1320 a
. 5200km
800-60 80 a
A cos c2-c b a
formula, cosineUsing
222
222
4
b
23. A satellite in space, an earth station and the
centre of earth all lie in the same plane. Let r be
radius of earth and R be distance from the
centre of earth to satellite. Let d be distance from
the earth station to the satellite. Let 30 be angle
of elevation from the earth station to satellite. If
the line segment connecting earth station and
satellite subtends angle α at the centre of earth,
