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WEIGHTAGE FOR CLASS ---XI
One Paper Three Hours Max Marks. 100
Units MarksI. SETS AND FUNCTIONS
II. ALGEBRA
III. COORDINATE GEOMETRY
IV. CALCULUS
V. MATHEMATICAL REASONING
VI. STATISTICS AND PROBABILITY
29
37
13
06
03
12 100
Fundamental Trigonometric Identities
Before we start to prove trigonometric identities, we see where the basic identities come from.
Recall the definitions of the reciprocal trigonometric functions, csc θ, sec θ and cot θ
from the trigonometric functions chapter:
After we revise the fundamental identities, we learn about: Proving trigonometric identities
Now, consider the following diagram where the point (x, y) defines an angle θ at the origin,
and the distance from the origin to the point is r units:
From the diagram, we can see that the ratios sin θ and cos θ are defined as:
and
Now, we use these results to find an important definition for tan θ:
Now, also so we can conclude that:
Also, for the values in the diagram, we can use Pythagoras' Theorem and obtain:
y2 + x2 = r2
Dividing through by r2 gives us:
so we obtain the important result:
sin2 θ + cos2 θ = 1
We now proceed to derive two other related formulas that can be used when proving trigonometric identities.
It is suggested that you remember how to find the identities, rather than try to memorise each one.
Dividing sin2 θ + cos2 θ = 1 through by cos2 θ gives us:
so
tan2 θ + 1 = sec2 θ
Dividing sin2 θ + cos2 θ = 1 through by sin2 θ gives us:
so
1 + cot2 θ = csc2 θ
Trigonometric Identities Summary
Proving Trigonometric Identities
Suggestions...
1. Learn well the formulas given above (or at least, know how to find them quickly).
The better you know the basic identities, the easier it will be to recognise what is going on in the problems.
2. Work on the most complex side and simplify it so that it has the same form as the simplest side.
Don't assume the identity to prove the identity.
This means don't work on both sides of the equals side and try to meet in the middle.
3. Start on one side and make it look like the other side.
4. Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only.
5. In most examples where you see power 2 (that is, 2 ), it will involve using the identity sin2 θ + cos2 θ = 1 (or one of the other 2 formulas that we derived above).
Using these suggestions, you can simplify and prove expressions involving trigonometric identities.
Prove that
sin y + sin y cot2 y = cosec y
Answer
FunctionAbbreviation
Description
Identities (using radians)
Sine sin
Cosine cos
Tangent tan (or tg)
Cotangent
cot (or ctg or ctn)
Secant sec
Cosecantcsc (or cosec)
Note that these values can easily be memorized in the form
but the angles are not equally spaced.
The values for 15°, 54° and 75° are slightly more complicated. [ by using formulas of sin(A-B),sin(A+B)
Similarly for cosine function & tan function.]
Special values in trigonometric functions There are some commonly used special values in
trigonometric functions, as shown in the following table.
Function
sin
0 1
cos 1 0
tan 0 1
cot 1 0
sec 1 2
csc 21
The sine and cosine functions graphed on the Cartesian plane.
For angles greater than 2π or less than −2π, simply continue to rotate
around the circle; sine and cosine are periodic functions with period 2π:
for any angle θ and any integer k.
The smallest positive period of a periodic function is called the
primitive period of the function.
The primitive period of the sine or cosine is a full circle, i.e.
2π radians or 360 degrees.
Figure 1
If Q(x,y) is the point on the circle where the string ends,
we may think of as being an angle by associating to it the central angle with vertex O(0,0) and sides passing through the points P and Q. If instead of wrapping a length s of string around the unit circle, we decide to wrap it around a circle of radius R, the angle (in radians) generated in the process will satisfy the following relation:
Observe that the length s of string gives the measure of the angle only when R=1.
As a matter of common practice and convenience, it is useful to measure angles in degrees, which are defined by partitioning one whole revolution into 360 equal parts, each of which is then called one degree. In this way, one whole revolution around the unit circle measures radians and also 360 degrees (or ), that is:
Each degree may be further subdivided into 60 parts, called minutes, and in turn each minute may be subdivided into another 60 parts, called seconds:
Angle sum identities
Sine
Illustration of the sum formula.
Draw the angles α and β. Place P on the line defined by α + β at unit distance from the origin.
Let PQ be a perpendicular from P to the line defined by the angle α. OQP is a right angle.
Let QA be a perpendicular from Q to the x axis, and PB be a perpendicular from P to the x axis. OAQ is a right angle.
Draw QR parallel to the x-axis. Now angle RPQ = α (because
or PQ/OP = Sin𝛃 or OQ/OP = Cos𝛃 (if OP ≠ 1)
, so
, so
Or = PBOP = PR+RB
OP = PROP . PQ
PQ + AQOP . OQ
OQ
(∵ RB = AQ , PQ & OQ are the hyp. )
= PRPQ . PQ
OP + AQOQ . OQ
OP .
By substituting − β for β and using Symmetry, we also get:
Cosine
Using the figure above,
OR PQ/OP = Sin𝛃 OR OQ/OP = Cos𝛃
, so
, so
Or = OBOP = OA−BA
OP = OAOP
. OQOQ - RQ
OP . PQPQ
(AB=RQ)
= OAOQ .OQ
OP - RQPQ . PQ
OP .
By substituting − β for β and using Symmetry, we also get:
Also, using the complementary angle formulae,
**Another simple "proof" can be given using Euler's formula known from complex analysis: Euler's formula is:
Although it is more precise to say that Euler's formula entails the trigonometric identities, it follows that for angles α and β we have:
ei(α + β) = cos(α + β) + isin(α + β)
Also using the following properties of exponential functions:
ei(α + β) = eiαeiβ = (cosα + isinα)(cosβ + isinβ)
Evaluating the product:
ei(α + β) = (cosαcosβ − sinαsinβ) + i(sinαcosβ + sinβcosα)
This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get:
cos(α + β) = cosαcosβ − sinαsinβ
sin(α + β) = sinαcosβ + sinβcosα ]
Tangent and cotangent
From the sine and cosine formulae, we get
Dividing both numerator and denominator by cos α cos β, we get
Similarly (using a division by sin α sin β), we get
Double-angle identities
From the angle sum identities, we get
and
The Pythagorean identities give the two alternative forms for the latter of these:
The angle sum identities also give
**It can also be proved using Eulers Formula
Mulitplying the exponent by two yields
But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields
It follows that
By multiplying we get
Because the imaginary and real parts have to be the same, we are left with the original identities
Half-angle identities
The two identities giving alternative forms for cos 2θ give these:
One must choose the sign of the square root properly—note that if 2π is added to θ the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore the correct sign to use depends on the value of θ.
tan function, we have
If we multiply the numerator and denominator inside the square root by (1 + cos θ), and do a little manipulation using the Pythagorean identities, we get
If instead we multiply the numerator and denominator by (1 - cos θ), we get
This also gives
Similar manipulations for the cot function give
Example. verify the identity
Answer. We have
which gives
But
and since
and , we get finally
Remark. In general it is good to check whether the given formula is correct. One way to do that is to substitute some numbers for the variables. For example, if we take a=b = 0, we get
or we may take . In this case we have
Example. Find the exact value of
Answer. We have
Hence, using the additions formulas for the cosine function we get
Since
we get
Example. Find the exact value for
Answer. We have
Since
we get
Finally we have
Remark. Using the addition formulas, we generate the following identities
Double-Angle and Half-Angle formulas are very useful. For example, rational functions of sine and cosine wil be very hard to integrate without these formulas. They are as follow
Example. Check the identities
Answer. We will check the first one. the second one is left to the reader as an exercise. We have
Hence
which implies
Many functions involving powers of sine and cosine are hard to integrate. The use of Double-Angle formulas help reduce the degree of difficulty.
Example. Write as an expression involving the trigonometric functions with their first power.
Answer. We have
Hence
Since , we get
or
Example: Verify the identity
Answer. We have
Using the Double-Angle formulas we get
Putting stuff together we get
From the Double-Angle formulas, one may generate easily the Half-Angle formulas
In particular, we have
Example. Use the Half-Angle formulas to find
Answer. Set . Then
Using the above formulas, we get
Since , then is a positive number. Therefore, we have
Same arguments lead to
Example. Check the identities
Answer. First note that
which falls from the identity . So we need to verify only one identity. For example, let us verify that
using the Half-Angle formulas, we get
which reduces to
Product and Sum Formulas
From the Addition Formulas, we derive the following trigonometric formulas (or identities)
Remark. It is clear that the third formula and the fourth
are identical (use the property to see it).
The above formulas are important whenever need rises to transform the product of sine and cosine into a sum. This is a very useful idea in techniques of integration.
Example. Express the product as a sum of trigonometric functions.
Answer. We have
which gives
Note that the above formulas may be used to transform a sum into a product via the identities
Example. Express as a product.
Answer. We have
Note that we used .
Example. Verify the formula
Answer. We have
and
Hence
which clearly implies
Example. Find the real number x such that
and
Answer. Many ways may be used to tackle this problem. Let us use the above formulas. We have
Hence
Since , the equation gives
and the equation gives . Therefore, the solutions to the equation
are
Example. Verify the identity
Answer. We have
Using the above formulas we get
Hence
which implies
Since , we get
TRIGONOMETRIC EQUATIONS
Example : Solve for x in the following equation.
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the tangent term.
= tan(±π
6 ) General solution:
X = nπ ± π6 ∀ n Є Z(integers)
Example : Solve for x in the following equation.
There are an infinite number of solutions to this problem. To solve for x, set the equation equal to zero and factor.
then
when
when , and when
when and
This is impossible because
The exact value solutions are and
Example : Solve for x in the following equation.
There are an infinite number of solutions to this problem.
Isolate the sine term. To do this, rewrite the left side of the equation in an equivalent factored form.
The product of two factors equals zero if at least one of the factors equals zeros. This means that
if or
We just transformed a difficult problem into two easier problems. To find the solutions to the original equation,
, we find the
solutions to the equations OR
Sinx = sin(-π/6) x = nπ + (-1)n (π/6) ∀ nЄ Z
OR
sinx = sin(π/2) x = nπ + (-1)n (π/2) ∀ nЄ Z.
Example : Solve for x in the following equation. (general solution)
There are an infinite number of solutions to this problem.
Cos(3x-1) = cos(π/2) 3x-1 = 2nπ ± π/2 3x = 2nπ ± π/2 +1 x = 1/3(2nπ ± π/2 +1) ∀ n Є Z.
1. Solve the trigonometric equation analytically
4 tan x − sec2 x = 0 (for 0 ≤ x < 2π)
Answer
4 tan x − sec2 x = 0
In 0 ≤ x < 2π, we need to find values of 2x such that 0 ≤ 2x < 4π.
So
So
or x = 0.2618, 1.309, 3.403, 4.451
2. Solve the trigonometric equation analytically for 0 ≤ x < 2π:
sin 2x cos x − cos 2x sin x = 0
Answer
We recognise the left hand side to be in the form:
sin(a − b) = sin a cos b − cos a sin b,
where a = 2x and b = x.
So
sin 2x cos x − cos 2x sin x
= sin(2x − x)
= sin x
Now, we know the solutions of sin x = 0 to be:
x = 0, π.
3. Solve the given trigonometric equation analytically and by graphical method (for 0 ≤ x < 2π):
sin 4x − cos 2x = 0
Answer
sin 4x − cos 2x = 0
2sin 2x cos 2x − cos 2x = 0
cos2x (2sin 2x - 1) = 0
EITHER
cos 2x = 0
OR
sin 2x = 1/2
Question Solve the equation tan 2θ − cot 2θ = 0 for 0 ≤ θ < 2π.
Answer
tan2 2θ = 1
tan 2θ = ± 1
Since 0 ≤ θ < 2π , we need to consider values of 2θ such that 0 ≤ 2θ < 4π. Hence, solving the above equation, we have:
So
Question Solve the equation for 0 ≤ θ < 2π.
Answer
By using the half angle formula for and then squaring both sides, we get:
= 1 + cos x
So we have:
2 cos2 x + 3 cos x + 1 = 0
(2 cos x + 1)(cos x + 1) = 0
Solving, we get
cos x = − 0.5 or cos x = − 1
Now gives .
However, on checking in the original equation, we note that
but
So the only solution for this part is
Also, cos x = − 1 gives x = π.
So the solutions for the equation are
GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS
1. If sinx = 0 ⇨ x = nπ, ∀ n є Z
2. If cosx = 0 ⇨ x = (2n+1) π/2, ∀ n Є Z
3. If tanx = 0 ⇨ x = nπ, n Є Z
4. If sinx = six ⇨ x = nπ + (-1)n𝛂, ∀ n Є Z
5. If cosx = cos ⇨ x = 2nπ ±𝛂, ∀ n Є Z
6. If tanx = tan ⇨ x = nπ+𝛂, ∀ n Є Z.
EXAMPLE: Solve the equation: sin3θ + cos2θ = 0
SOLUTION: cos2θ = - sin3θ ⇨ cos2θ = cos (π2 +3θ)
⇨ 2θ = 2nπ± (π2
+3θ) , ∀ nЄZ
-θ = nπ+ π2
and 5θ = nπ- π2 , ∀ nєz.
Question –1 If cos(A+B)=4/5 , sin(A-B)=5/13 and A,B lie between 0 and π/4 , prove that
tan 2A = 56/33.
Answer: Since A-B Є (-π/4 , π/4) and A+B Є (0,π/2) both are positive sin(A+B)=3/5 , cos(A-B)=12/13
tan(A+B)=3/4 , tan(A-B) = 5/12 then
tan2A = tan(A+B+A-B) =
34+ 5
12
1−34
×5
12
=56/33
Question – 2 If cos (A-B)+cos(B-C)+cos(C-A) = -3/2 , Prove that
CosA+cosB+cosC = sinA+sinB+sinC = 0
Answer : From given result we get
2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosAcosC+2sinAsinC +3 =0
2cosAcosB+2sinAsinB+2cosBcosC+2sinBsinC+2cosAcosC+2sinAsinC + sin2A+cos2A+sin2B+cos2B
+sin2C+cos2C = 0. ( 3=1+1+1 and 1 can be written as sin2x+cos2x)
(sin2A+sin2B+sin2C+ 2sinAsinB+2sinBsinC+2sinAsinC)
+(cos2A+cos2B+cos2C +2cosAcosB+2cosBcosC+2cosAcosC) = 0
(sinA+sinB+sinC)2+ (cosA+cosB+cosC)2 = 0.
Question – 3 Prove that (1+cos π8 ) (1+ cos
3 π8 ) (1+
cos 5 π8 ) (1+ cos
7 π8 ) =
18 .
Answer : cos 7 π8 = cos( π -
π8 ) = - cos
π8 , cos
5 π8
= cos ( π - 3 π8 ) = - cos
3 π8
L.H.S. (1+cos π8 ) (1+ cos
3 π8 ) (1+ cos
5 π8
) (1+ cos 7 π8 )
= (1 – cos2 π8 ) (1 – cos2
3 π8 ) = sin2 π
8
. sin2 3 π8
= 14 (2sin2 π
8 ). (2sin2 3 π8 ) =
14 (1 -
cos π4 ) ( 1 - cos
3 π4 ) =
18 .
Question – 4 Prove that cos 2 π15 cos
4 π15 cos
8 π15 cos
14 π15 =
116
.
Answer : L.H.S. ⇨ - cos 2 π15 cos
4 π15 cos
8 π15 cos
π15
[∵cos 14 π15 = cos(π -
π15 ) ]
= - [ sin 16 A
24 SinA ] , where A= π15
[∵ all angels are in G.P. , short-cut Method]
= - [sin 16
π15
24 sinπ
15
] = - sin(π+ π
15)
16 sinπ
15
=
sinπ
15
16sinπ
15
= 116
. [ ∵ sin (π + π15
) lies in 3rd quadrant]
OR
L.H.S. ⇨ - 1
2sinπ15
(2 sin π15 cos
π15 ) cos
2 π15
cos 4 π15 cos
8 π15
= - 1
2× 2sinπ
15 ( 2sin
2 π15 cos
2 π15 )cos
4 π15
cos 8 π15
= - 1
2× 2× 2sinπ
15 ( 2sin
4 π15 cos
4 π15 )cos
8 π15
= - 1
2× 2× 2× 2 sinπ
15 (2 sin
8 π15 .cos
8 π15 )
, now you can apply above result.
Similarly you can prove cosAcos2Acos4Acos8A = sin16A/16sinA.
Question – 5 (i) Prove that sin200 sin400 sin600 sin800 = 3
16
Answer : L.H.S. sin200 sin400 sin600 sin800
(√3/2¿
⇨ √ 32× 2 (2sin200 sin400 sin800)
⇨ √ 34 [(cos200 – cos600) sin800 ]
⇨ √ 34 [cos200 . sin800– cos600. sin800 ]
(1/2)
⇨ √ 38 [(sin1000+sin600 – sin800 ] = 3
16 ( ∵ sin1000 lies in 2nd quadrant) [sin(1800 - 800)] = sin800
(ii) Prove that: cosπ7 cos
2 π7 cos
4 π7 = -
18 .
[Hint: let x = π7 , then
12 sinx (2sinx cosx cos2x cos4x)]
Sin2x (iii) Prove that: tan200 tan400 tan800 = tan600.[hint: L.H.S. sin 200sin 400sin 800
cos200 cos 400 cos800 solve as above method.] Question – 6 Solve : 2sinx + √3 cosx = 1+ sinx
Answer : sinx + √3 cosx = 1
√ 32 cosx +
12 sinx =
12 [ dividing by
√a2+b2 ,where a = √3 and b = 1]
cosπ6 . cosx + sin
π6 .sinx = cos
π3
⇨ cos(x - π6 ) = cos
π3 ⇨ x - π
6 = 2nπ ±π3 ∀ n Z (integers)Є
⇨ x = 2nπ ±π3 + π
6 = 2nπ + π2 , 2nπ - π
6 ∀ n Z Є(integers).Question – 7 Solve: 3cos2x - 2 √3 sinx .cosx – 3sin2x = 0. Answer: 3 cos2x - 3 √3 sinx .cosx +√3 sinx .cosx – 3sin2x = 0 ⇨ 3cosx ( cosx - √3 sinx) + √3 sinx (cosx - √3 sinx) = 0 ⇨ (3cosx + √3 sinx) (cosx - √3 sinx) = 0 ⇨ (3cosx + √3 sinx) = 0 or (cosx - √3 sinx) = 0
⇨ tanx = - √3 = tan(-π3 ) or tanx = 1
√ 3 =tan(π6 )
⇨ x = nπ+(- π3 ) ∀ n Z Є(integers) or x = nπ + π
6
Question – 8 If 𝛂, are the acute angles and cos2𝛂 = 3 cos2 β−13−cos2 β , show that tan 𝛂 =
√ 2 tan𝛃. Answer: According to required result , we have to convert given part into tangent function By using cos2𝛂 = 1−tan ² α
1+tan ² α ∴ we will get 1−tan ² α
1+tan ² α = 3( 1−tan ² β
1+tan ² β)−1
3−1−tan ² β1+ tan ² β
= 3−3 tan ² β−1− tan ² β3+3 tan ² β−1+ tan ² β
= 1−2 tan ² β1+2 tan ² β
By C & D 1− tan ² α+1+ tan ² α
(1−tan ² α)−(1+ta n2 α) = 1−2 tan ² β+1+2 tan ² β
(1−2 tan ² β)−(1+2 tan2 β) ⇨ 2−2 tan ² α =
2−4 tan ² β
2
⇨ tan ² α = 2 tan ² β ⇨ tan 𝛂 = √ 2 tan𝛃.
Question – 9 Prove that sinA−sin 5 A+sin 9 A−sin 13 A
cosA−cos 5 A−cos 9 A+cos13 A = cot4A.
Answer : (sin 9 A+sinA )−¿¿ = 2 sin5 Acos4 A−2 sin 9 Acos 4 A
−[−2sin 5 Asin4 A ]+[−2 sin 9 Asin 4 A]
[∵ sinA+sinB = 2sin(A+B)/2. cos(A-B)/2 & cosA-cosB = - 2sin(A+B)/2.sin(A-B)/2 ]
= 2cos 4 A (sin 5 A−sin 9 A)2 sin 4 A(sin 5 A−sin 9 A)
= cot 4A.
Question – 10 Prove that √2+√2+√2+2 cos 8θ = 2COSθ. Answer: √2+√2+√2(1+cos8 θ) = √2+√2+√2×2cos ²4 θ
√2+√2(1+cos 4 θ) = √2+√2× 2cos ²2θ = √2(1+cos2θ) = 2cosθ. [By using 1+cos2 θ= 2cos2θ] Question – 11 Find the general solution of the following equation:
4sinxcosx+2sinx+2cosx+1 = 0
Answer: Above equation can be written as (4sinxcosx+2sinx) + (2cosx+1 ) = 0
⇨ 2sinx (2cosx+1) + (2cosx+1) = 0
⇨ (2sinx+1) (2cosx+1) = 0
⇨ (2sinx+1) = 0 or 2cosx+1 = 0
⇨ sinx = -1/2 or cosx = -1/2
⇨ sinx = sin(π + π6 )
or cosx = cos (π - π3 )
⇨ x = nπ +(-1)n 7 π6
or x = 2nπ ± 2 π3 , ∀ n Є Z (Integers).
Question – 12 If cosx = -1213 and π <x <3 π
2 , find the
value of sin3x and cos3x.
Answer: Since x lies in 3rd quadrant ∴ sinx is negative.
Sinx = – √1−cos ² x = - 513 then sin3x = 3sinx
– sin3x = -20352197
Cos3x = 4cos3x – 3cosx = - 8282197 [by
putting the values of sinx & cosx]
Question – 13 (i) If cos(A+B) sin(C-D) = cos(A-B) sin(C+D) , then show that
tanA tanB tanC + tanD = 0
(ii) If sinθ = n sin(θ+2𝛂), prove that
tan(θ+𝛂) = 1+n1−n tan𝛂.
Answer: We can write above given result as cos (A+B)cos( A−B) = sin(C+D)
sin(C−D)
By C & D cos ( A+B )+cos ( A−B)cos ( A+B )−cos( A−B) =
sin (C+D )+sin(C−D)sin (C+D )−sin(C−D)
cosA . cosB−sinA . sinB = sinC .cosD
cosC . sinD
-cotA.cotB = tanC.cotD
- 1tanA . 1
tanB = tanC. 1tanD
⇨ - tanD = tanA tanB tanC ⇨ tanA tanB tanC + tanD = 0 .
(ii) sin(θ+2α )sinθ = 1
n , by C & D sin (θ+2α )+sinθsin (θ+2α)−sinθ = 1+n
1−n
⇨ 2 sin (θ+α )cosα2 sinα cos (θ+α) = 1+n
1−n ⇨ tan(θ+𝛂) = 1+n1−n tan𝛂.
Question – 14 Prove that
(i) cos 520+cos 680+cos 1720 = 0
(ii) sinA .sin(600 – A).sin(600 + A) = 14 sin3A.
(it lies in 2nd quad.)
Answer: (i) L.H.S. cos 520+(cos 680+cos 1720) = cos520 + 2 cos1200 cos520
= cos520 + 2 ׿ ) cos520 [∵ cos1200 = cos(1800-600)]
= 0
(ii) L.H.S. sinA.[sin(600 – A).sin(600 + A)] = sinA [sin2 600 – sin2 A]
=
sinA [34 - sin2 A] = 1
4 [3sinA – 4sin3A]
= 1
4 sin3A.
[We know that sin2A-sin2B = sin(A+B)sin(A-B) & cos2A – sin2B = cos(A+B)cos(A-B)]
Question – 15 If sinx + siny = a and cosx + cosy = b, find
the values (i) tan( x− y2
) (ii) tan¿.
Answer: sinx + siny = a ⇨ 2sin( x+ y2
¿ cos¿) =
a ..............1
cosx + cosy = b ⇨ 2cos( x+ y2
¿ cos¿) =
b ..............2
(i) By squaring & adding above results, we get
4 cos² ¿) [sin²( x+ y2
¿ + cos²( x+ y2
¿] = a2+b2
( 1 )
Sec2( x− y2
) = 4a ²+b ² ⇨ tan2( x− y
2) = 4
a ²+b ² - 1 ⇨ tan
( x− y2
) = √ 4−a ²−b ²a ²+b ²
.
(ii) dividing 1 by 2, we get tan( x+ y2
) = ab .
Question – 16 Prove that
(i) tan700 = 2 tan500 + tan200 .
(ii) Prove that: tan300+ tan150+ tan300. tan150 =1.
(iii) cos2 A + cos2 B – 2cosAcosBcos(A+B) = sin2 (A+B).
Solution: (i) We have tan700 = tan(500 + 200)
= tan 50 °+ tan 20 °1−tan50 ° . tan 20 °
⇨ tan700 [1−tan 50 ° . tan 20 °] = tan50 °+ tan 20°
⇨ tan700 - tan700. tan 50 ° . tan 20° = tan50 °+ tan 20°
⇨ tan700 – tan(900-200). tan 50 ° . tan 20° = tan50 °+ tan 20°
(cot200 = 1tan 20 ° )
⇨ tan700 – tan500 = tan50 °+ tan 20°
⇨ tan700 = 2 tan500 + tan200.
(ii) [hint: take tan450 = tan(300+150)
(iii) L.H.S. cos2 A + cos2 B – (2cosAcosB)cos(A+B) = cos2 A + cos2 B – (cos(A+B)+cos(A-B))cos(A+B)
cos2 A + cos2 B – [cos2(A+B)+cos(A-B)cos(A+B)]
cos2 A + cos2 B – [cos2(A+B)+cos2(A) – sin2(B]
cos2 A + cos2 B - cos2(A+B)-cos2(A) + sin2(B]
1 - cos2(A+B) = sin2 (A+B).
Question – 17 If x+y = π4 , prove that (i) (1+tanx)
(1+tany) = 2 (ii) (cotx – 1)(coty – 1) = 2.
Answer: (i) tan(x+y) = tanπ4 ⇨ tanx+tany
1−tanx . tany = 1
⇨ tanx+tany+tanx.tany = 1
⇨ (1+ tanx) + tany(1 + tanx) = 1+1
⇨ (1+tanx)(1+tany) = 2
(ii) Similarly for second part by using cot(x+y) = cotx . coty−1cotx+coty .
Question – 18 Prove that (i) tan1890 = cos36 °−sin 36 °cos36 °+sin 36 °
(ii) Find the value of tan220 30’.
Answer: (i) we can take both sides to prove above result
L.H.S. tan1890 = tan(1800 +90) = tan9° =
tan(45°-36°) = tan 45 °+ tan36 °1−tan 45 ° . tan36 °
= 1−tan36 °
1+tan 36 ° = cos36 °−sin 36 °cos36 °+sin 36 ° .
(ii) Let x= 220 30’ then 2x = 450
We know that tanx = √ 1−cos2 x1+cos2 x ⇨ tanx
= √ 1−cos 45°1+cos45 °
= √ 1− 1√ 2
1+ 1√ 2
= √ (√2−1 )(√2−1)(√2+1 )(√2−1)
tan220 30’ = √2−1 [∵ it lies in 1st quad.]
Question: If tanx+tany = a and cotx+coty = b, prove that 1/a - 1/b = cot(x+y).
Answer: L.H.S. = 1tanx+tany - 1
coty+cotx = cosxcosy−sinxsinysinxcosy−cosxsiny
{after simplification}.
Question – 19 Prove that cos2 x + cos2 (x+π3 ) + cos2 (x-π
3
) = 32 .
Answer: L.H.S. cos2 x + cos2 (x+π3 ) +[1 - sin2 (x-π
3 ) ]
⇨ 1+ cos2 x + [ cos2 (x+π3 ) - sin2 (x-π
3 ) ]
⇨ 1+ cos2 x + [ cos (x+π3
+x− π3
¿ ) cos(x+π3
-x+π3 ) ] ⇨ 1+ cos2 x + [cos (2x) .cos(2 π
3 ) ]
( ∵cos(π - π3
¿ it lies in 2nd quad.)
⇨ 1+ cos2 x + [cos (2x) ׿ )] ⇨ 2+2 cos² x−cos2 x2 =
2+2cos² x−(2cos ² x−1)2
= 2+2 cos² x−2 cos² x+12 = 3
2 .
Question – 20 Prove that
(a) sin3x + sin3(2 π3 +x) + sin3(4 π
3 +x) = - 34
sin3x.
[Hint: Use sin3A = 3sinA – 4sin3A ⇨ 4sin3A = 3sinA - sin3A ⇨ sin3A = ¼[3sinA - sin3A]]
(b) If tanx + tan(x+π3 ) + tan¿+x) = 3 , then
show that tan3x = 1. [ use formula of tan(x+y)]
(c) Find in degrees and radians the angle subtended b/w the hour hand and the minute hand
Of a clock at half past three. [answer is 750 , 5π/12]
(d) 1tan 3 A−tanA - 1
cot 3 A−cotA = cot 2A. [Hint
put cot3A = 1/tan3A & cotA = 1/tanA then take l.c.m.
and use formula tan(A-B).]
(e) If α , 𝛃 are the distinct roots of acosθ +
bsinθ = c, prove that sin(𝛂+𝛃) = 2 aba ²+b ² .
Solution of (e) If α , 𝛃 are the distinct roots of acosθ + bsinθ = c, then
acos𝛂 + bsin𝛂 = c & acos𝛃 + bsin𝛃 = c
By subtracting , we get a(cos𝛂 – cos𝛃) + b(sin𝛂 – sin𝛃) = 0 ⇨ a(cos𝛂 – cos𝛃) = b(sin𝛂 – sin𝛃) ⇨ 2a sin α+β2 sin α−β
2 = 2b cosα+β2 sin
α−β2
⇨ tanα+β2 = b
a ⇨ sin(𝛂+𝛃) = 2 tan
α+β2
1+ tan ²α+β
2
⇨
sin(𝛂+𝛃) = 2aba ²+b ² .
Question: If sin2A = k sin2B, prove that tan( A+B)tan (A−B) =
k+1k−1
[ Hint by c & d sin 2 A+sin 2 B
sin 2 A−sin 2 B = k+1k−1 , use formula of sinx+siny= 2sin( x+ y
2¿ cos¿)]
Question: If cos(x+2A) = n cosx, show that cotA = n+11−n tan(x+A). Question: Prove that tan(∝−β ¿ = sin 2 β
5−cos 2 β , if 2tan∝ = 3tanβ. [Hint: L.H.S. = tanα−tanβ1+ tanα . tanβ , put the value of tan∝ and simplify it]. Question: If sinx + siny = a and cosx + cosy = b,find the
value of cos(x-y).
Answer: squaring and adding above results, we will get cos(x-y) = ½[a2+b2 – 2]
Question: Show that 1sin 10° - √ 3
cos10 ° =4
[Hint: 2 [ 1/2sin 10° - √3/2
cos10 ° ] = 2 [ sin 30°sin 10° - cos30 °
cos10 ° ]
Question: Prove that : 1+sinx−cosx1+sinx+cosx = tan(x/2)
[ Hint: use 1 – cosx = 2sin2x/2 , 1+cosx = 2cos2x/2 and sinx = 2sinx/2.cosx/2]
** Question Solve: √2 sec∅ + tan∅ = 1
Solution √2 + sin∅ = cos∅ ⇨ cos∅ - sin∅ = √2 dividing by √a2+b2 =√2 ∵ a=1,b=1 (cos∅ - sin∅ )/ √2 = 1 ⇨ cos( /4)п cos∅ - sin( /4) п sin∅ =1 ⇨ cos(∅+ /4) = cos0п 0 ∅+ /4 = 2nп п±0, n∈Z ⇨ ∅ = 2n –п /4.п
** Question: If tan2A = 2tan2B + 1, prove that cos2A + sin2B =0.
Answer: L.H.S. 1−tan ² A1+ tan ² B + sin2B =
−2 tan ² B
2(1+tan2 B) + sin2B , by
putting above result and simplify it.
** Question: Find the maximum and minimum values of sinx+cosx.
Answer: maximum value of (asinx+bcosx) = √a2+b2 = √ 2
Minimum value of (asinx+bcosx) = - √a2+b2 = - √ 2
Or
√ 2[ 1√ 2 sinx + 1
√ 2 cosx] = √ 2 sin(x+п/4), as -1 ≤
sin(x+п/4) ≤1 ∀ x
**Question: Find the minimum value of 3cosx+4sinx+8. [ answer is 3 as above result]
**Question: ∀ x in (0,п/2), show that cos(sinx) > sin(cosx).
Answer: п/2> √ 2 [∵п/2 = 22/7=1.57 and √ 2 =1.4]
We know that п/2 > √ 2 ≥ sinx+cosx ⇨ п/2 - sinx > cosx ⇨ cos(sinx) > sin(cosx).
** Question: If A = cos2x + sin4x ∀ x, prove that ¾ ≤ A ≤ 1
Answer: A = cos2x + sin2x. sin2x ≤ cos2x + sin2x ⇨ A ≤ 1 , A = (1 - sin2x) + sin4x = [sin2x – (½)]2 + (¾) ≥ ¾.
** Question: (i) find the greatest value of sinx.cosx [ ½.(2sinx.cosx) ≤ ½]
(ii) If sinx and cosx are the roots of ax2 – bx+c = 0, then a2 – b2 +2ac = 0.
[ Hint: sum and product of roots ⇨ (sinx+cosx)2 = 1+2(c/a)]
** Question : If f(x) = cos2x+sec2x , then find which is correct f(x)<1, f(x)=1, 2< f(x)<1, f(x)≥2.
[A.M. ≥ G.M.⇨ f(x)/2= (cos2x+sec2x)/2 ≥√ (cos2x.sec2x) , ANSWER IS f(x)≥2].
** Question: solve 3sin 2x+2 cos ² x+31−sin2 x +2sin² x = 28.
[hint: use 2cos2x = 1+ cos2x , 2sin2x = 1 – cos2x and let sin2x+cos2x = y
Above equation becomes 31+ y+32− y = 28. Then put t = 3y ⇨ 3t2 – 28t +9=0 ⇨ t=1/3 , 9
When y=-1 then sin2x+cos2x = -1 ⇨ cosx = 0 or tanx = -1
When y = 2 then sin2x+cos2x =2 ⇨ 1√ 2 sin2x+ 1
√ 2 cos2x = √ 2 ⇨ sin(2x+π /4 ¿=√ 2 >1 which is not possible