TRIGONOMETRY
Definition
The word ‘trigonometry’ is derived from two Greek words; trigonon = triangle and metron = to measure. The literal meaning of trigonometry is “ to measure a triangle.” At present it is defined as that branch of mathematics which deals with angles, whether of a triangle or any other figure. In trigonometry, we study about the three angles of a triangle and also about its sides.
Reciprocal Ratios : Consecant, secant
And contangent ratios are the reciprocal P
Rations of sine, cosine, and tangent ratios
Respectively.
Angle : An angle is the amount of
Revolution made by a straight line about O X
One of its extreme in a plane from one
Position to another.
Symbolic Terms
In a right angle triangle ABC
AB = (Perpendicular) = p
BC = (Base) = b
AC = (Hypotenuse) = h
I. h2 = p2 + b2 h = II. p2 = h2 - b2 p =
III. b2 = h2 - p2 b =
Trigonometric Ratios
Since the three sides of a right angle triangle have six possible ratios, besides sine, cosine and the tangent of an angle , there are three other trigonometric ratios, namely cosecant, secant and cotangent of the angle . Trigonometric ratios are as follows:
I. Sine = = , written as sin
II. Cosine = = , written as cos
III. Tangent = = , written as tan
IV. Contangent = = , written as cot
V. Secant = = , written as sec
VI. Cosecant = = , written as cosec
Important Formulae
1. sin = 2. cosec =
2. sin × cosec = 1 4. cos =
5. sec = 6. cos × sec = 1
7. tan = 8. cot =
9. tan × cot = 1 10. tan =
11. tan = 12. cot =
13. cot = 14. sin2 + cos2 = 1
15. sin2 = 1 - cos2 16. 1 - sin2 = cos2
17. 1 + tan2 = sec2 18. sec2 - tan2 = 1
19. tan2 = sec2 – 1 20. 1 + cot2 = cosec2
21. cot2 = cosec2 - 1 22. cosec2 - cot2 = 1
23. cos = 24. sin =
25. = 1
Memorable Facts
I. In case of a right angled triangle the trigonometric function can be defined as under.
Let ∆ABC be the right angled triangle in which
AC = h = hypotenuse
AB = p = perpendicular
BC = b = base and ACB =
sin = =
cos = =
tan = =
cot = =
sec = =
cosec = =
Relation between p, b, h
According to Phthogoras theorem
(hypotenuse)2 = (perpendicular)2 + (base)2
h2 = p2 + b2 h =
p2 = h2 - b2 p =
b2 = h2 - p2 b =
II. Trigonometric ratio can be calculated for both acute and obtuse angle, but here we study for acute angle only.
In trigonometry, Greek symbols are used which are given here
= alpha β = beta γ = gamma
, = delta λ = lambda ψ = psi
= theta ϗ = kappa Ф = phi
= pi = = 180o
Fundamental relationships between trigonometric functions with proof:
1. sin × cosec = 1 L.H.S. = left hand side
R.H.S. = right hand side
Soln. L.H.S. = sin × cosec = × = 1 = R.H.S.
(a) sin = (b) cosec =
2. sec × cos = 1
Soln. L.H.S. = sec × cos = × = 1 = R.H.S.
(a) sec = (b) cos =
3. tan × cot = 1
Soln. L.H.S. = tan × cot = × = 1 = R.H.S.
(b) tan = (b) cot =
4. = tan
Soln. L.H.S. = = = × = = tan = R.H.S.
5. = cot
Soln. L.H.S. = = = × = = cot = R.H.S.
6. sin2 + cos2 = 1
Soln. L.H.S. = sin2 + cos2 = + = = = 1 R.H.S.
(a) sin2 = 1 – cos2 sin = (b) cos2 = 1 – sin2 cos =
6. sec2 - tan2 = 1
Soln. L.H.S. = sec2 - tan2
= ( )2 – ( )2 = - = = = 1 R.H.S.
(a) sec2 = 1 + tan2 sec = (b) tan2 = sec2 – 1 tan =
7. cosec2 - cot2 = 1
Soln. L.H.S. = cosec2 - cot2
= ( )2 – ( )2 = - = = = 1 R.H.S.
(a) cosec2 = 1 + cot2 cosec = (b) cot2 = cosec2 – 1 cot =
Note :
(a) Numerical value of sin and cos is less than or equal to unity.
Ex. sin 1; cos 1
(b) Numerical value of cosec and sec is equal to or greater than unity.
Ex. cosec ≥ 1; sec ≥ 1
(c ) Any numerical value can be positive for tan and cot .
Ex. - < tan < ; - < cot <
Q. If cot = , then find the value of
(i) cos (ii) sin
Soln. Here cot = = = k (Let) b = 12k , p = 5k
h = =
= = = 13k
(i) cos = = = ; (ii) sin = = = Ans.
Trigonometric ratios
1. Trigonometric ratios for 45o
Let ∆ABC be the right angled isosceles triangle in which A = 90o
B = C = 45o
Suppose AC = x, so due to isosceles triangle AB = AC = x
For right angled ∆ABC
BC2 = AB2 + AC2
or, h2 = p2 + b2
or, h2 = x2 + x2
or, h = x = BC
Now, sin 45o = sin ACB = = =
cos 45o = cos ACB = = =
tan 45o = tan ACB = = = 1
cot 45o = cot ACB = = = 1
sec 45o = sec ACB = = =
cosec 45o = cosec ACB = = =
2. Trigonometric ratios for 30o and 60o :
Let ABC be an equilateral triangle
Then ABC = ACB = BAC = 60o
or, A = B = C = 60o
Now, perpendicular AD is drawn from point A to straight line BC. Due to equilateral ∆, AD bisect BC and A.
BAD = CAD = 30o
In right angled ABD
h2 = p2 + b2
p2 = h2 - b2
⇒ AD2 = AB2 – BD2
or, AD2 = (2x)2 – x2
or, AD2 = 4x2 – x2 = 3x2
AD = x
(a) Trigonometric ratios for 30o
sin 30o = sin BAD = = =
cos 30o = cos BAD = = =
tan 30o = tan BAD = = =
cot 30o = cot BAD = = =
sec 30o = sec BAD = = =
cosec 30o = cosec BAD = = = 2
Note : For A = , BD = perpendicular,
AD = base, AB = hypotenuse
(b) Trigonometric ratios for 60o
sin 60o = sin ABD = = =
cos 60o = cos ABD = = =
tan 60o = tan ABD = = =
cot 60o = cot ABD = = =
sec 60o = sec ABD = = = 2
cosec 60o = cosec ABD = = =
Note : For B = , AD = perpendicular,
BD = base, AB = hypotenuse
(c) Conversion of one system into another system :
180P
oP = radian 1P
oP =
30 = × 30 = radian 45 = × 45 = radian
60 = × 60 = radian 90 = × 90 = radian
Note : radian = 180˚ = 200P
gP = 2 rt. Angles
or, = =
(d) Remember it :
ratios = 0 = 30 = 45 = 60 = 90 sin
= 0 = = = = 1
cos = 1 = = = = 0
(e) Arranging in a regular order :
ratios when increases
sin , tan , sec increases
cos , cot , cosec decreases
If A > B, then
sin A > sin B, cos A < cos B, tan A > tan B
cot A < cot B, sec A > sec B, cosec A < cosec B
(f) Trigonometric ratios of standard angles :
ratios = 0 30˚= 45˚= 60˚= 90˚= sin 0
1
cosec 2 1
cos 1 0
sec 1 2
tan 0 1
cot 1 0
Q. Find the value of tanP
2P60˚ cosecP
2P30˚ tan45˚
Soln. tan 60 = , cosec 30˚ = 2, tan 45˚ = 1
Now from question = ( )P
2 P(2)P
2P 1 = 3 × 4 × 1 = 12 Ans.
3. Trigonometric ratios of (90˚ - ).
Let ABC be the right angled triangle in which BCA = 90˚.
If BAC = , then CBA = (90˚ - )
Now, sin (90˚ - ) = sin CBA =
sin (90˚ - ) = cos
Again, cos (90˚ - ) = cos CBA =
and sin = sin BAC =
cos (90˚ - ) = sin
Again, tan (90˚ - ) = tan CBA =
and cot = cot BAC =
tan (90˚ - ) = cot
Similarly cosec (90˚ - ) = sec
sec (90˚ - ) = cosec
cot (90˚ - ) = tan
Definition relating to height & distances :
Horizontal plane : The plane which is parallel to earth plane is called horizontal plane.
Horizontal line : The straight line which is parallel to earth plane is called horizontal line.
Line of sight : The straight line which is passing through the observer (eye) and object is called line of sight.
Angle of elevation : Angle between the horizontal line and line of sight is called angle of elevation when object is above the observer.
Angle of depression : Angle between the horizontal line and line of sight is called angle of depression when object is below the observer.
Formulae for Higher Mathematics(Intermediate)
Measurement of Angles :
(a) Sexagesimal or degree system
1 right angle = 90˚ 1˚ = 60’ (60 minute)
1’ = 60” (60 second)
(b) Sentesimal grade system
1 right angle = 100 g (g grade) 1 grade = 100’ (100 minute)
1 minute = 100’ (100 second)
(c.) Circular system or radian measure
1 right angle = radian
1 radian = = 57˚17’43”(approx)
Similarly 2 right angle = radian
Conversion one system into another system
radians = 180˚ = 200g = 2rt. Angles
or, = (1 radian = 57˚17’ & 1˚ = 0.01746 radian)
Ratio of circumference to diameter of a circle is called :
= or, C = x d; = = 3.1416 (approx)
Radian : When an arc of a circle makes equal value to that of radius, then angle made by arc is called radian.
angle of radian = or, =
If length of arc = radius, then = 1˚
Trigonometrical rationals :
Quadrants : If we take two mutually perpendicular fixed lines XOX’ and YOY’ in a plane, the plane is divided into four equal parts, each part is known as a quadrant.
Y’
2nd quadrant 1st quadrant
(- , +) (+ , +)
X’ o X
(- , -) (+ , -)
3rd quadrant 4th quadrant
1st quadrant (0˚ - 90˚) = all are positive.
2nd quadrant (90˚ - 180˚) = only sine and cosecant are positive.
3rd quadrant (180˚ - 270˚) = only tangent and contangent are positive.
4th quadrant (270˚ - 360˚) = only cosine and secant are positive.
Y’ 90˚
only sin, cosec all are positive
are positive
X’ 180˚ o 0˚ X
only tan, cot only cos, sec
are positive are positive
Y 270˚
(a) -1 sin 1; for all value of
-1 cos 1; for all value of
tan any real number
cot any real number
cosec 1 or cosec -1 for all value of
1 sec or, sec -1 for all value of
(b) If > , then
sin > sin , when -90 < , < 90
sin < sin , when 90 < , < 270
cos > cos , when -180 < , < 0
cos < cos , when 0 < , < 180
tan > tan , when , are in same quadrant.
(c )sin(- ) = - sin cos(- ) = cos
tan(- ) = - tan cot(- ) = - cot
sec(- ) = - sec cosec(- ) = - cosec
sin(90 - ) = cos cos(90 - ) = sin
tan(90 - ) = cot cot(90 - ) = tan
sec(90 - ) = cosec cosec(90 - ) = sec
sin(90 + ) = cos cos(90 + ) = -sin
tan(90 + ) = -cot cot(90 + ) = -tan
sec(90 + ) = -cosec cosec(90 + ) = sec
sin(180 - ) = sin cos(180 - ) = -cos
tan(180 - ) = -tan cot(180 - ) = -cot
sec(180 - ) = -sec cosec(180 - ) = cosec
sin(180 + ) = -sin cos(180 + ) = -cos
tan(180 + ) = tan cot(180 + ) = cot
sec(180 + ) = -sec cosec(180 + ) = -cosec
sin(360 + ) = sin cos(360 + ) = cos
tan(360 + ) = tan cot(360 + ) = cot
sec(360 + ) = sec cosec(360 + ) = cosec
sin(2n + ) = sin cos(2n + ) = cos
tan(2n + ) = tan cot(2n + ) = cot
sec(2n + ) = sec cosec(2n + ) = cosec
sin(360 - ) = -sin cos(360 - ) = cos
tan(360 - ) = -tan cot(360 - ) = -cot
sec(360 - ) = sec cosec(360 ) =-cosec
sin(2n - ) = -sin cos(2n - ) = cos
tan(2n - ) = -tan cot(2n - ) = -cot
sec(2n - ) = sec cosec(2n - ) = -cosec
In this case, the sign of the resulting ratio is taken according to the quadrant in which the angle lies. Above results are valid for acute as well as obtuse angles.
Q. (i) Find the value of tan 495˚.
Soln. tan (360˚ + 135˚) = tan 135˚ = tan (180˚ - 45˚)
= -tan 45˚ = -1 Ans.
(ii) sin (-1332˚) = -sin 1332˚ = -sin(3 x 360˚ + 252˚)
= -sin 252˚ = -sin (180˚ + 72˚) = -(-sin 72˚)
= sin 72˚ = sin (90˚ - 18˚) = + cos 18˚ Ans.
Trigonometric ratios of compound angles :
If A and B are two angles, positive, zero or negative, we have
Sum of two angles in sine, cosine, tangent
(a) sin (A + B) = sin A cos B + cos A sin B (b) cos (A + B) = cos A cos B - sin A sin B
(c) tan (A + B) =
Difference Formulae :
(d) sin (A - B) = sin A cos B - cos A sin B
Ratio = 120˚; = 135˚; = 150˚; = = 180˚
sin 0
cos - -
- -1
tan - - 1 - 0
sec - 2 - - -1
cosec 2
cot - - 1 -
(e) cos (A - B) = cos A cos B + sin A sin B
(f) tan (A - B) =
Other Formulae :
(g) cot (A + B) =
(h) cot (A - B) =
(i) sin (A + B) sin (A - B) = sin2A - sin2B = cos2B - cos2A (j) cos (A + B) cos (A - B) = cos2A - sin2B = cos2B - sin2A (k) sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C
- tan A tan B tan C) (l) cos (A + B + C) = cos A cos B cos C (1 - tan B tan C - tan C tan A
- tan A tan B)
(m) tan (A + B + C) =
(n) a cos + b sin = cos ( – ) where cos =
= sin ( + ) where sin =
Q. Prove that : tan 5A – tan 3A – tan 2A = tan 5A tan 3A tan 2A.
Sol. 3A + 2A = 5A
taking tan in both side of above equation
or, tan (3A + 2A) = tan 5A
or, = tan 5A
or, tan 3A + tan 2A = tan 5A – tan 5A tan 3A tan 2A
or, tan 5A - tan 3A - tan 2A = tan 5A tan 3A tan 2A
Q. Prove that + + = 0
L.H.S. = + +
= + +
= - + - + -
= tan B – tan C + tan C – tan A + tan A – tan B = 0 R.H.S. Proved.
Transformation formulae : Sum and Product :
(a) sin (A + B) + sin (A – B) = 2 sin A cos B (b) sin (A + B) - sin (A – B) = 2 cos A sin B (c) cos (A + B) + cos (A – B) = 2 cos A cos B (d) cos (A - B) - cos (A + B) = 2 sin A sin B (e) sin (A + B) sin (A – B) = sin2 A – sin2 B (f) cos (A + B) cos (A – B) = cos2 A – sin2 B
Product formulae :
If A + B = C and A – B = D, then A = , B =
(a) sin C + sin D = 2 sin cos
(b) sin C - sin D = 2 cos sin
(c) cos C + cos D = 2 cos cos
(d) cos C - cos D = 2 sin sin
Prove that : cos 4 cos 2 – cos 6 cos 12 = sin 10 sin 8
Soln. L.H.S. = cos 4 cos 2 – cos 6 cos 12
= (2 cos 4 cos 2 – 2 cos 6 cos 12 )
= [{cos (4 + 2 ) + cos (4 - 2 )}] – [{cos (12 + 6 ) + cos (12 - 6 )}]
= [cos 6 + cos 2 – cos 18 – cos 6 ]
= [ cos 2 – cos 18 ]
= x 2 sin sin
= sin 10 sin 8 = R.H.S. Proved.
Q. Find the value of sin 10˚ sin 50˚ sin 60˚ sin 70˚.
Soln. sin 60˚ sin 10˚ (2 sin 70˚ sin 50˚)
= x sin 10˚ ( cos 20˚ cos 120˚)
= x (2 sin 10˚ cos 20˚ - 2 sin 10˚ cos 120˚)
= [ ( sin 30˚ - sin 10˚) – 2 (sin 10˚) (- )
= [ sin 30˚ - sin 10˚ + sin 10˚ ]
= x = = Ans.
Multiple Angles :
(a) sin 2A = 2 sin A cos A =
(b) cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A =
(c) tan 2A =
(d) 1 + cos 2A = 2 cos2 A (e) 1 – cos 2A = 2 sin2 A
(f) tan2 A =
(g) sin 3A = 3 sin A – 4 sin3 A (h) cos 3A = 4 cos3 A – 3 cos A
(i) tan 3A =
(j) cot 2A =
(k) cot 3A =
(l) sin3 A =
4 sin3 A = 3 sin A – sin 3A
(m) 4 cos3 A = 3 cos A + cos 3A
cos3 A =
Q. Prove that : cosec 10˚ - sec 10˚ = 4
Soln. L.H.S. = - =
=
= 2 x 2 x
= = = 4 = R.H.S. Proved.
Submultiple Angles :
(a) sin A = 2 sin cos =
(b) cos A = cos2 - sin2 = 2 cos2 – 1 = 1 – 2 sin2 =
(c) tan A =
(d) 1 + cos A = 2 cos2
(e) 1 - cos A = 2 sin2
(f) = tan2
(g) sin 18˚ =
(h) cos 18˚ =
(i) cos 36˚ = ( + 1)
(j) sin 36˚ =
(k) sin A = 3 sin - 4 sin3
(l) cos A = 4 cos3 - 3 cos
(m)tan A =
(n) sin 15˚ =
(o) cos 15˚ =
(p) tan 15˚ = 2 -
(q) sin 75˚ = cos 15˚ =
(r) cos 75˚ = sin 15˚ =
(s) tan 75˚ = 2 + (t) cot 75˚ = 2 -
(u) sin 22 = cos 67 =
(v) cos 22 = sin 67 =
(w) tan 22 = cot 67 = - 1
(x) cos =
(y) sin =
(z) tan =
Q. Prove that : = cot
Soln. L.H.S. = =
= = = cot = R.H.S. Proved.
Trigonometrical Identities
If A + B + C = 180˚, then
A + B = 180˚ - C; B + C = 180˚ - A; A + C = 180˚ - B.
(i) sin (A + B) = sin (180˚ - C) = sin C
sin (B + C) = sin (180˚ - A) = sin A
sin (A + C) = sin (180˚ - B) = sin B
(ii) cos (A + B) = cos (180˚ - C) = -cos C
cos (B + C) = cos (180˚ - A) = -cos A
cos (A + C) = cos (180˚ - B) = -cos B
(iii) tan (A + B) = tan (180˚ - C) = -tan C tan (B + C) = tan (180˚ - A) = -tan A
tan (A + C) = tan (180˚ - B) = -tan B
Similarly if = + + , then + = -
Sin ( + ) = sin ( - ) = cos
Trigonometrical Equation with principal solution : The solution of equations which lie in the interval, ( 0, 2 ) are called the principal solutions. Solutions are always given in radians,
1. (a) If sin x = sin , where 0˚ < < 90˚
then, x = , 180˚ -
(b) If sin x = sin , where - 90˚ < < 0˚
then, x = (360˚ + ), (180˚ - )
2. If cos x = cos , where 0˚ < < 180˚, then, x = , (360˚ - ) 3. If tan x = tan , where 0˚ < < 180˚, then, x = , (180˚ + )
General Solution : The solutions giving all the possible angles, satisfying the equations are called “General Solutions.”
General Expressions :
(i) If sin = 0, then = n (ii) If cos = 0, then = (2n + 1)
(iii) If tan = 0, then = n (iv) If sin = sin , then = n + (-1)n (v) If cos = cos , then = 2n (vi) If tan = tan , then = n + (where n is any integer)
Quadratic Equation : The general values of satisfying the equations sin2 = sin2 , cos2 = cos2 and tan2 = tan2 is = (n ).
Table of trigonometrical ratio :
ratio 120˚ 225˚ 240˚ 270˚ 300˚ 315˚ 330˚ 360˚
sin - - - -1 - - - 0
cos - - - 0 1
tan 1 - - 1 - 0
cot 1 0 - - 1 -
sec - - - 2 2
cosec - 2 - - 1 - - - 2
Properties of Triangles
1. sine formula ( In triangle ABC)
= = = 2R
or, = = =
where R is the radius of circle circumscribed around the triangle.
2. cosine formula ( In triangle ABC)
a2 = b2 + c2 – 2bc cos A or, cos A =
b2 = a2 + b2 – 2ac cos B or, cos B =
c2 = a2 + b2 – 2ab cos C or, cos C =
3. Projection formula (In triangle ABC)
a = b cos C + c cos B
b = c cos A + a cos C
c = a cos B + b cos A
4. Half angle formula (BRIGG’s formula)
(i) sin = – –
sin = – –
sin = – – [ where s = ]
(ii) cos = –
cos = –
cos = –
(iii) tan = – –
tan = – –
tan = – –
5. Area formulae for triangle
(a) = ab sin C
(b) = bc sin A
(c) = ca sin B
6. (a) =
(b) = ; R =
(c ) sin A = , sin B = , sin C =
7. (a) sin A =
(b) sin B =
(c ) sin C =
8. (a) tan =
(b) tan =
(c ) tan =
9. (a) cot =
(b) cot =
(c ) cot =
10. (a) tan = cot
(b) tan = cot
(c ) tan = cot
Q. In any ABC, prove that + + =
L.H.S. = + +
= + +
= ( –
= = R.H.S. Proved.
Inverse Circular Functions
1. (i) sin-1 (sin ) = (ii) cos-1 (cos ) =
(iii) tan-1 (tan ) = (iv) cosec-1 (cosec ) =
(v)sec-1 (sec ) = (vi) cot-1 (cot ) =
2. (i) sin (sin-1x) = x (ii) cos (cos-1x) = x
(iii) tan (tan-1x) = x (iv) cosec (cosec-1x) = x
(v) sec (sec-1x) = x (vi) cot (cot-1x) = x
3. (i) sin-1 x = cosec-1 ( ), cosec-1 x = sin-1 ( )
(ii) cos-1 x = sec-1 ( ), sec-1 x = cos-1 ( )
(iii) tan-1 x = cot-1 ( ), cot-1 x = tan-1 ( )
4. (i) sin-1 x + cos-1 x = (ii) tan-1 x + cot-1 x =
(iii) sec-1 x + cosec-1 x =
5. (i) tan-1x + tan-1y = tan-1
(ii) tan-1x - tan-1y = tan-1
(iii) sin-1x + sin-1y = sin-1{ x + y }
(iv) sin-1x - sin-1y = sin-1{ x - y }
(v)cos-1x + cos-1y = cos-1 { xy – }
(vi)cos-1x - cos-1y = cos-1 { xy + }
(vii)tan-1x + tan-1y + tan-1 = tan-1
6. (i) 2 sin-1x = sin-1 (2x )
(ii) 2 cos-1x = cos-1(2 - 1)
(iii)2 tan-1 x = tan-1( )= sin-1( )
7. (i) 3 sin-1x = sin-1(3x – 4x3)
(ii) 3 cos-1x = cos-1 (4x3 – 3x)
8. (i) sin-1x = cos-1
= tan-1 = cot-1
= sec-1 ( ) = cosec-1 ( )
9. (ii) cos-1x = sin-1 = tan-1
= ( ) = cot-1 ( )
= sec-1 ( ) = cosec-1 ( )
Demaivre’s Theorem
1. (cos + isin )n = cos n + isin n 2. (cos + isin )-n = cos n - isin n 3. (cos - isin )n = cos n - isin n 4. (cos - isin )-n = cos n - isin n
5. = (cos + isin )-1 = cos - isin
6. sin n = n cosn-1 sin –
cosn - 3 sin3 +
cosn - 5 sin5 +….
7. cos n = cosn – cosn - 2 sin2
+ x cosn – 1 sin4 +……
8. tan n =
=
=
9. cosn and sinn in terms of z.
If z = cos + i sin
= cos - i sin
z + = 2 cos z - = 2 i sin