Class XI CBSE-Mathematics Trigonometric Functions...Class–XI–CBSE-Mathematics Trigonometric...

Class–XI–CBSE-Mathematics Trigonometric Functions

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CBSE NCERT Solutions for Class 11 Mathematics Chapter 03

Back of Chapter Questions

1. Find the radian measures corresponding to the following degree measures: (i)25o

(ii)−47o30′

(iii)240o

(iv)520o

Solution:

(i) Step1: Given degree is 25o

It is known that 180o = πradian

Therefore, 25o =𝜋

180× 25radian

=5𝜋

36 radian

Overall Hint: Convert the degree measures to radians by using the formula Angle in

degree = 𝜋

180× 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑛𝑔𝑙𝑒 𝑖𝑛 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 radian

(ii) Step1: Given degree is −47o30’

−47o30′ = −471

2degree [∵ 1o = 60′]

= − 95

2

o

As180o = 𝜋radian

⇒ − 95

2

o=

𝜋

180× (

−95

2)radian= (

−19

72) 𝜋radian

Therefore, −47o30’ = (−19

72) 𝜋radian


degree = 𝜋


(iii) Step1: Given degree is240o

It is known that 180o = 𝜋radian

⸫240o =𝜋

180× 240radian



=4𝜋

3 radian


degree = 𝜋


a. Step1: Given degree is520o

As 180o = 𝜋radian

Therefore, 520o =𝜋

180× 520radian=

26𝜋

9radian


degree = 𝜋


2. Find the degree measures corresponding to the following radian measures(Useπ =22

7)

(i)11

16

(ii)−4

(iii)5π

3

(iv)7π

6

Solution:

(i) Step1:

Given radian is11

16

As 𝜋radian= 180o

⇒ 11

16radian=

180

π×

11

16degree=

45×11

𝜋×4degree

=45×11×7

22×4degree

=315

8degree

= 393

8degree

Step2:

= 39o +3×60

8Minutes [1o = 60′]



= 39o + 22’ +1

2minutes

=39o22′30” [1’ = 60”]

Overall Hint: Convert the given radian angle into degrees by using the formula Rad × 180/π = Deg

(ii) Step1: Given radian is −4

As 𝜋radian= 180𝑜

−4radian=180

π× (−4)radian=

180×7×(−4)

22degree

=−2520

11degree= −229

1

11degree

= −229o +1×60

11minutes [1o = 60′]

Step2:

= −229o + 5’ +5

11minutes

= −229o + 5’ +5×60

11 minutes

= −229o5′27" [1’ = 60”]

Overall Hint: Convert the given radian angle into degrees by using the formula Rad ×

180/π = Deg

(iii) Step1:

Given radian is 5𝜋

3

As πradian= 180o

5𝜋

3radian=

180

π×

5𝜋

3radian

= 300o Overall Hint: Convert the given radian angle into degrees by using the formula Rad × 180/π = Deg

(iv) Step1:

Given radian is7𝜋

6

As 𝜋radian= 180o



7𝜋

6radian=

180

π×

7𝜋

6radian= 210o

Overall Hint: Convert the given radian angle into degrees by using the formula Rad × 180/π = Deg

3. A wheel makes 360revolutions in one minute. Through how many radians does it turn in one second?

Solution:

Step1:

Given,

Numberofrevolutionsmadebythewheelin1minute= 360

Therefore, Numberofrevolutionmadebythewheelin1second=360

60= 6

For one completerevolution,thewheel willturnanangleof2πradian.

Therefore,in6completerevolutions,itwillturnanangleof= 6 × 2π = 12π radian

Hence,inonesecond,thewheelturnsanangleof12πradian.

Overall Hint: Convert the revolutions per minute to radians per second by using the formula 1 rpm = 2(pi)/60 = pi/30 radians per second

4. Find the degree measure of the angle subtended at the Centre of a circle of

radius100 cm by an arc of length22 cm (Useπ =22

7)

Solution:

Step1:

Given,

radiusofcircle= 100 cm,

lengthofarc= 22 cm.

Consider theanglesubtendedbythearcattheCentreofthecircleasθ.



Weknow, θ = arc

radius

Step2:

θ =22

100radian=

180

π×

22

100degree=

180×7×22

22×100degree

=126

10degree

= 123

5degree

= 123×60

5 degree [1o = 60′]

= 12o36′

Therefore,therequiredangleis12o36’. Overall DL: M

Overall Hint: Find the angle subtended at the center of the circle by using the formula

θ = arc

radius then find the angle measure in degree by using θ =

180×7×22

22×100degree

5. In a circle of diameter40 cm,the length of a chord is20 cm.Find the length of minor arc of the chord.

Solution:

Step1:

Step1:

Step2:

Given, Diameter of the circle= 40 cm

So, Radius of the circle=40

2cm = 20 cm

Let AB be a chord of the circle.



InΔOAB, OA = OB =radiusofcircle= 20 cm

Also, AB = 20 cm

Therefore ΔOAB is equilateral triangle.

⸫θ = 60o = 𝜋

3radian

Step3:

We know that θ = arc

radius

Arc= θ ×Radius

= 𝜋

3× 20 cm

=20𝜋

3cm

Therefore, the length of the minor arc of the chord is 20𝜋

3cm.

Overall Hint: Draw a rough figure making use of the given data then find the measure

of the angle in radians. Also find the arc length using the formula Arc= θ ×Radius

6. If in two circles,arcs of the same length subtend angles60oand75oat the Centre,find the ratio of their radii.

Solution:

Step1:

Let the angle formed by the arc of first circle θ1 = 60o = 60 × (π

180)radian=

60𝜋

180

Let the angle formed by the arc of second circle θ2 = 75o = 75 × (π

180) =

75π

180

Let the radius of first circle be 𝑟1 and radius of second circle be 𝑟2

It is known that, θ = arc

radius

Step2:

Radius=arc

θ



𝑟1

𝑟2 =

𝑎𝑟𝑐

𝜃1𝑎𝑟𝑐

𝜃2

= 𝜃2

𝜃1

r1

r2=

75π

18060π

180

=75

60=

5

4

Therefore, the ratio of their radii is 5: 4 Overall Hint:First find the angles formed by the arcs of the two circles in radians then

make use of the formula 𝑟1

𝑟2 =

𝑎𝑟𝑐

𝜃1𝑎𝑟𝑐

𝜃2

= 𝜃2

𝜃1 to find ratio of their radii.

7. Find the angle in radian through which a pendulum swings if its length is75 cm and the tip describes an arc of length.

(i)10 cm

(ii)15 cm

(iii)21 cm

Solution:

We know that in a circle of radius 𝑟, if an arc of length 𝑙 unit subtends an angle θ radian

at the Centre, then 𝜃 = 𝑙

𝑟

Given that 𝑟 = 75 cm (i) Step1:

Here,𝑙 = 10 cm

𝜃 = 𝑙

𝑟

𝜃 = 10

75radian=

2

15radian

Overall Hint: Find the angle using formula

𝜃 = 𝑙

𝑟 radian

(ii) Step1:

Here,𝑙 = 15 cm

𝜃 = 𝑙

𝑟

⇒ θ = 15

75radian=

1

5radian




𝜃 = 𝑙

𝑟 radian

(iii) Step1:

Here,𝑙 = 21 cm

𝜃 = 𝑙

𝑟

θ = 21

75radian

=7

25radian


𝜃 = 𝑙

𝑟 radian

Exercise3.2

Find the values of other five trigonometric functions in Exercise 1 to 5.

1. cos 𝑥 = −1

2, 𝑥 lies in third quadrant.

Solution:

Step1:

Given, cos 𝑥 =1

2

Therefore, sec 𝑥 = 1

cos 𝑥=

1

−1

2

= −2

We know that sin2𝑥 + cos2𝑥 = 1

Step2:

⇒ sin2𝑥 = 1 − cos2𝑥

⇒ sin2𝑥 = 1 − (−1

2)

2

⇒ sin2𝑥 = 1 −1

4

⇒ sin2𝑥 =3

4

Therefore, sin 𝑥 = ±√3

2



The value of sin𝑥 is negative as 𝑥 lies in third quadrant. So, sin 𝑥 = −√3

2

Step3:

We know that cosec 𝑥 =1

sin𝑥

⇒ cosec 𝑥 = −2

√3

We know that tan 𝑥 =sin𝑥

cos𝑥=

−√3

2

−1

2

Therefore, tan 𝑥 = √3

We know that cot 𝑥 =1

tan𝑥

Therefore, cot𝑥 =1

√3

Overall Hint: As cos x is given we can find sec x as its reciprocal then we will find sin x using 1 – cos^2(x) and then cosec x as its reciprocal the cot x and tan x can be found using the ratio of cos x and sin x

2. sin 𝑥 =3

5, 𝑥Liesinsecondquadrant.

Solution:

Step1:

Given, sin 𝑥=3

5

cosec 𝑥 =1

sin𝑥

⇒ cosec 𝑥 =5

3

We know that, sin2𝑥 + cos2𝑥 = 1

Step2:

⇒ cos2𝑥 = 1 − sin2𝑥

⇒ 𝑐𝑜𝑠2𝑥 = 1 − (3

5)

2

⇒ cos2𝑥 = 1 −9

25



⇒ cos2𝑥 =16

25

⇒ cos 𝑥 = ±4

5

The value of cos is negative as 𝑥 lies in second quadrant. So, cos 𝑥 = −4

5

Step3:

We know that, sec 𝑥 =1

cos𝑥

⇒ sec 𝑥 = −5

4

We know that, tan 𝑥 =sin𝑥

cos𝑥

=

3

5

−4

5

= −3

4

We know that, cot 𝑥 =1

tan𝑥= −

4

3

Overall Hint: As sin x is given we can find cosec x as its reciprocal then we will find cos x using 1 – sin^2(x) and then sec x as its reciprocal the cot x and tan x can be found using the ratio of cos x and sin x

3. cot 𝑥 =3

4, 𝑥 lies in third quadrant.

Solution:

Step1:

Given, cot 𝑥 =3

4

We know that, tan 𝑥 =1

cot𝑥

⇒ tan 𝑥 =4

3

We know that cosec2𝑥 = 1 + cot2𝑥

⇒ cosec2𝑥 = 1 + (3

4)

2

⇒ cosec2𝑥 = 1 +9

16=

25

16



⇒ cosec 𝑥 = ±√25

16

Therefore, cosec 𝑥 = ±5

4

Step2:

But x lies in third quadrant so cosec𝑥 will be negative.

⇒ cosec 𝑥 = −5

4

We know that sin 𝑥 =1

cosec𝑥

sin 𝑥 = −4

5

Also,sin2𝑥 + cos2𝑥 = 1


⇒ cos2𝑥 = 1 − (−4

5)

2

⇒ cos2𝑥 = 1 −16

25

⇒ cos2𝑥 =9

25

⇒ cos 𝑥 = ±3

5

Step3:

But 𝑥 lies in third quadrant,so value of cos𝑥 is negative.

Therefore, cos 𝑥 = −3

5

sec 𝑥 = 1

cos 𝑥= −

5

3

Overall Hint: As cot x is given we can find tan x as its reciprocal then we will find cosec x using 1 + cot^2(x) and then sin x as its reciprocal the cos x and sec x can be found using the 1- sin^(2)x and the reciprocal of cos x will be sec x

4. sec 𝑥 =13

5, 𝑥 lies in fourth quadrant.



Solution:

Step1:

Given, sec 𝑥 = 13

5

⇒ cos 𝑥 =1

sec𝑥=

5

13

We know that sin2𝑥 + cos2𝑥 = 1

⇒ sin2𝑥 = 1 − cos2𝑥

⇒ sin2𝑥 = 1 − (5

13)

2

⇒ sin2𝑥 =144

169

⇒ sin 𝑥 = ±12

13

Step2:

But 𝑥 lie in a fourth quadrant, so value of sin 𝑥 is negative.

Therefore, sin 𝑥 = −12

13

Also,cosec 𝑥 =1

sin𝑥= −

13

12

tan 𝑥 =sin 𝑥

cos 𝑥=

−12

135

13

= −12

5

cot 𝑥 =1

tan𝑥

Therefore, cot 𝑥 = −5

12

Overall Hint: As sec x is given we can find cos x as its reciprocal then we will find sin x using 1 – cos^2(x) and then cosec x as its reciprocal the cot x and tan x can be found using the ratio of cos x and sin x and vice versa

5. tan x = −5

12, x in second quadrant.

Solution:

Step1:



Given, tan 𝑥 = −5

12

cot 𝑥 =1

tan𝑥

⇒ cot 𝑥 = −12

5

We know that cosec2𝑥 = 1 + cot2𝑥

⇒ cosec2𝑥 = 1 + (−12

5)

2

⇒ cosec2𝑥 = 1 +144

25=

169

25

⇒ cosec𝑥 = ±√169

25

⇒ cosec𝑥 = ±13

5

Step2:

But 𝑥 lies in second quadrant, so cosec𝑥 is positive.

⸫cosec𝑥 =13

5

We know that sin 𝑥 =1

cosec 𝑥

sin 𝑥 =5

13

Also,sin2𝑥 + cos2𝑥 = 1


⇒ cos2𝑥 = 1 − (5

13)

2

⇒ cos2𝑥 = 1 −25

169

⇒ cos2𝑥 =144

169

⇒ cos 𝑥 = ±12

13



Step3:

But 𝑥 lies in these con quadrant, so value of cos𝑥 is negative.

Therefore, cos 𝑥 = −12

13

We know that, sec 𝑥 = 1

cos 𝑥

Therefore, sec 𝑥 = − 13

12

Overall Hint: As tan x is given we can find cot x as its reciprocal then we will find cosec x using 1 + cot^2(x) and then sin x as its reciprocal the cos x can be found using 1 – sin^2(x) and sec x as its reciprocal

Find the values of the trigonometric functions in Exercises 6 to 10

6. sin 765o

Solution:

Step1:

sin 765o can be written as sin(2 × 360o + 45o)

= sin 45o[∵ sin is positive in first quadrant]

=1

√2

Therefore, sin 765o = 1

√2

Overall Hint: Write sin 765o as sin (2*360 + 45) using Sin(2𝜋+ x) = sin x we get sin

765 as 1

√2

7. cosec (−1410o)

Solution:

Step1:

Given,cosec(−1410o)

= −cosec(1410o)

= − cosec(4 × 360o − 30o)

= −cosec(−30o)



= cosec(30o)

= 2

Overall Hint: Write cosec (−1410o) as cosec (4 × 360o − 30o) using cosec

(2n𝜋+x) = Cosec we get cosec (−1410o) as 2

8. tan19π

3

Solution:

Step1:

Given, tan19𝜋

3

tan19𝜋

3= tan (6𝜋 +

𝜋

3)

= tan𝜋

3 [We know that value of tan 𝑥 repeats after an interval of 𝜋 or 180o]

= √3

Therefore, tan19π

3= √3

Overall Hint: Write tan19π

3 as tan(6𝜋 +

𝜋

3) and using tan(2n𝜋 + 𝑥) = tan x we get

tan19π

3= √3

9. sin (−11π

3)

Solution:

Step1:

Given, sin (−11𝜋

3)

sin (−11𝜋

3) = − sin (

11𝜋

3) [∵sin is negative in fourth quadrant]

= − sin (4𝜋 −𝜋

3)

= − [− sin (𝜋

3)] [∵sin is negative in fourth quadrant]



=sin (𝜋

3)

=√3

2

Overall Hint: Write sin (−11π

3) as - sin (

11𝜋

3) = - sin (4𝜋 −

𝜋

3) then using sin (2n𝜋 - x)

= -sin x we get sin (−11π

3) as

√3

2

10. cot (−15π

4)

Solution:

Step1:

Given, cot (−15𝜋

4)

cot (−15𝜋

4) = − cot (

15𝜋

4) [∵cot is negative in fourth quadrant]

= − cot (4𝜋 −𝜋

4)

= − [− cot (𝜋

4)] [∵cot is negative in fourth quadrant]

= cot (𝜋

4)

= 1

Overall Hint: Write cot (−15π

4) as - cot (

15𝜋

4) then as -- cot (4𝜋 −

𝜋

4) using cot (2n𝜋-

x) = cot x we get cot (−15π

4) as 1

Exercise:3.3

1. sin2 π

6+ cos2 π

3− tan2 π

4= −

1

2

Solution:

L.H.Ssin2 𝜋

6+ cos2 𝜋

3− tan2 𝜋

4

= (1

2)

2

+ (1

2)

2

− (1)2

= 1

4+

1

4−1



=1

2− 1

= −1

2

=R.H.S

Hence LHS = RHS.

Overall Hint: Substitute the values of sinπ

6cos

π

3, tan

π

4 in the LHS . This will be equal

to what’s given in the RHS hence proved

2. Prove that : 2 sin2 π

6+ cosec2 7π

6cos2 π

3=

3

2

Solution:

L.H.S= 2 sin2 π

6+ cosec2 7π

6cos2 π

3

= 2 (1

2)

2

+ cosec2 (𝜋 +𝜋

6) (

1

2)

2

= 2 × 1

4+ (−cosec

𝜋

6)

2

(1

4)

=1

2+ (−2)2 (

1

4)

=1

2+ 1

=3

2

=R.H.S

Hence LHS = RHS.

Overall Hint: Substitute the values of sinπ

6, cosec

7π

6, cos

π

3 in the LHS . This will be

equal to what’s given in the RHS

3. Prove that cot2 π

6+ cosec5 π

6+ 3 tan2 π

6= 6

Solution:

L.H.S= cot2 𝜋

6+ cosec

5𝜋

6+ 3tan2 𝜋

6



= (√3)2

+ cosec (𝜋 −𝜋

6) + 3 (

1

√3)

2

= 3 + cosec (𝜋

6) + 3 ×

1

3

= 3 + 2 + 1

= 6

=R.H.S

Hence, LHS = RHS.

Overall Hint: Substitute the values of cotπ

6, cosec

π

6, tan

π



4. Prove that 2 sin2 3π

4+ 2 cos2 π

4+ 2 sec2 π

3= 10

Solution:

GIVEN, L.H.S= 2sin2 3𝜋

4+ 2cos2 𝜋

4+ 2sec2 𝜋

3

2sin2 (𝜋 −𝜋

4) + 2 (

1

√2)

2

+ 2(2)2

= 2sin2 (𝜋

4) + 2 ×

1

2+ 2 × 4

= 2 (1

√2)

2

+ 1 + 8

= 1 + 1 + 8

= 10

=R.H.S

Hence LHS = RHS.

Overall Hint: Substitute the values of sin3π

4, cos

π

4, sec

π



5. Find the value of: (i)sin 75o

(ii)tan 15o



Solution:

(i) Step1: sin75𝑜 = sin(45𝑜 + 30𝑜)

= sin 45ocos 30o + cos 45osin 30o [∵ sin(𝑋 + 𝑌) = sin 𝑋 cos𝑌 +cos𝑋 sin𝑌]

=1

√2×

√3

2+

1

√2×

1

2

=√3

2√2+

1

2√2

=√3 + 1

2√2

Overall Hint: The value of sin 75 can be found using sin (60+15) with the help of

formula sin(𝑋 + 𝑌) = sin 𝑋 cos𝑌 + cos𝑋 sin𝑌 by taking 60 and 15 as X and Y respectively

(ii) Step1:

tan15𝑜 = tan(45𝑜 − 30𝑜)

=tan45o−tan30𝑜

1+tan45otan30o [∵ tan(𝑥 − 𝑦) =tan 𝑥−tan 𝑦

1+tan 𝑥 tan 𝑦]

=1 −

1

√3

1 + 1 ×1

√3

=√3 − 1

√3 + 1

= (√3 − 1)

2

(√3 − 1) (√3 + 1)=

3 + 1 − 2√3

(√3)2

− 12=

4 − 2√3

3 − 1

= 2 − √3 Overall Hint: The value of tan 15 can be found using tan(45-30) with the help of

formula tan(𝑥 − 𝑦) =tan 𝑥−tan 𝑦

1+tan 𝑥 tan 𝑦 by taking 45 and 30 as x and y respectively

6. Prove the following :cos (π

4− x) cos (

π

4− y) − sin (

π

4− x) sin (

π

4− y) = sin (x + y)

Solution:



Step1:

L.H.S= cos (𝜋

4− 𝑥) cos (

𝜋

4− 𝑦) − sin (

𝜋

4− 𝑥) sin (

𝜋

4− 𝑦)

=1

2[2cos (

𝜋

4− 𝑥) cos (

𝜋

4− 𝑦)] +

1

2[−2sin (

𝜋

4− 𝑥) sin (

𝜋

4− 𝑦)]

=1

2[cos {(

𝜋

4− 𝑥) + (

𝜋

4− 𝑦)}] +

1

2[cos {(

𝜋

4− 𝑥) − (

𝜋

4− 𝑦)}]

+1

2[cos {(

𝜋

4− 𝑥) + (

𝜋

4− 𝑦)}] −

1

2[cos {(

𝜋

4− 𝑥) − (

𝜋

4− 𝑦)}]

[∵ 2cos𝐴cos𝐵 = cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵) and −2sin𝐴 sin𝐵 = cos(𝐴 + 𝐵) −cos(𝐴 − 𝐵)]

Step2:

=1

2× 2cos (

𝜋

2− (𝑥 + 𝑦))

= cos [𝜋

2− (𝑥 + 𝑦)]

= sin (𝑥 + 𝑦)

=R.H.S

Hence RHS = LHS Overall Hint: By multiplying the LHS throughout with 2 and dividing it with ½ and

using the identity 2cos𝐴cos𝐵 = cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵) and −2sin𝐴 sin𝐵 = cos(𝐴 + 𝐵) − cos(𝐴 − 𝐵) We can prove the LHS and RHS as equal

7. Prove that:tan(

π

4+x)

tan(π

4−x)

= (1+tanx

1−tanx)

2

Solution:

Step1:

We know that,

tan(𝑥 + 𝑦) =tan 𝑥+tan 𝑦

1−tan 𝑥 .tan 𝑦 andtan(𝑥 − 𝑦) =

tan 𝑥−tan 𝑦

1+tan 𝑥.tan 𝑦

Given, L.H.S=tan(

𝜋

4+𝑥)

tan(𝜋

4−𝑥)



=

tan𝜋

4+tan 𝑥

1−tan𝜋

4tan 𝑥

tan𝜋

4−tan 𝑥

1+tan𝜋

4tan 𝑥

=

1+tan 𝑥

1−tan 𝑥1−tan 𝑥

1+tan 𝑥

= (1 + tan 𝑥

1 − tan 𝑥)

2

=R.H.S

Hence, LHS = RHS

Overall Hint:Use the identities tan(𝑥 + 𝑦) =tan 𝑥+tan 𝑦

1−tan 𝑥 .tan 𝑦 andtan(𝑥 − 𝑦) =

tan 𝑥−tan 𝑦

1+tan 𝑥.tan 𝑦

In the numerator and denominator of The LHS to prove the RHS

8. Prove that cos(π+𝑥)cos(−𝑥)

sin(π−𝑥)cos(π

2+𝑥)

= cot2𝑥

Solution:

Step1:

Given, L.H.S=cos(𝜋+𝑥)cos (−𝑥)

sin(𝜋−𝑥)cos(𝜋

2+𝑥)

=(−cos 𝑥)( cos𝑥)

(sin 𝑥) (−sin𝑥)

=cos2𝑥

sin2𝑥

= cot2𝑥

=R.H.S

Hence LHS = RHS

Overall Hint: Cos(𝜋+x) = - cos x and cos (-x) = cos x and sin(𝜋 − 𝑥) = sinx, cos(𝜋

2+

𝑥) – sin x in the numerator and denominator of the LHS to prove the RHS



9. cos (3π

2+ 𝑥) cos(2π + 𝑥) [cot (

3π

2− 𝑥) + cot(2π + 𝑥)] = 1

Solution:

Step1:

Given, L.H.S.= cos (3𝜋

2+ 𝑥) cos (2𝜋 + 𝑥) [cot (

3𝜋

2− 𝑥) + cot (2𝜋 + 𝑥)]

= sin 𝑥 cos 𝑥 [tan 𝑥 + cot 𝑥]

= sin 𝑥 cos 𝑥 [𝑠𝑖𝑛 𝑥

cos 𝑥+

cos 𝑥

sin 𝑥]

= sin 𝑥 cos 𝑥 [sin2 𝑥 + cos2 𝑥

cos 𝑥 sin 𝑥]

= sin2𝑥 + cos2𝑥

= 1

=R.H.S.

Hence LHS = RHS.

Overall Hint: Use Cos(2𝑛+1𝜋

2+ 𝑥) = sin x , cos(2𝑛𝜋 + 𝑥 ) = cos x and [cot (

3𝜋

2− 𝑥) +

cot (2𝜋 + 𝑥)] = tan 𝑥 + cot 𝑥 in the LHS to prove the RHS

10. Prove that sin(n + 1) 𝑥 sin(n + 2) 𝑥 + cos(n + 1) 𝑥 cos(n + 2) 𝑥 = cos 𝑥

Solution:

Step1:

Given, L.H.S.= sin(𝑛 + 1) 𝑥 sin(𝑛 + 2)𝑥 + cos(𝑛 + 1) 𝑥 cos(𝑛 + 2)𝑥

=1

2(2 sin(𝑛 + 1) 𝑥 sin(𝑛 + 2)𝑥 + 2 cos(𝑛 + 1) 𝑥 cos(𝑛 + 2)𝑥)

=1

2[cos{(𝑛 + 1)𝑥 − (𝑛 + 2)𝑥} − cos{(𝑛 + 1)𝑥 + (𝑛 + 2)𝑥}]

+1

2[cos{(𝑛 + 1)𝑥 + (𝑛 + 2)𝑥} + cos{(𝑛 + 1)𝑥 − (𝑛 + 2)𝑥}]

=1

2× 2cos {(𝑛 + 1)𝑥 − (𝑛 + 2)𝑥}

= cos(−𝑥)



= cos 𝑥

=R.H.S

Hence LHS = RHS Overall Hint: Multiply the LHS of the equation throughout by 2 and divide with ½ then

use the formula cos(A − B) = cos A cos B + sin A sin B to prove the RHS

11. Prove that cos (3π

4+ 𝑥) − cos (

3π

4− 𝑥) = −√2 sin 𝑥

Solution:

Step1:

L.H.S= cos (3𝜋

4+ 𝑥) − cos (

3𝜋

4− 𝑥)

We know that,cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵

2) sin (

𝐴−𝐵

2)

So, L.H.S.= −2sin {(

3𝜋

4+𝑥)+(

3𝜋

4−𝑥)

2} sin {

(3𝜋

4+𝑥)−(

3𝜋

4−𝑥)

2}

= −2sin (3𝜋

4) sin𝑥

= −2sin (𝜋 −𝜋

4) sin𝑥

= −2sin (𝜋

4) sin𝑥

= −2 ×1

√2sin𝑥

= −√2sin𝑥

=R.H.S

Hence Proved

Overall Hint: Use the formula cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵

2) sin (

𝐴−𝐵

2) in the LHS to

prove the RHS. Also use sin (𝜋 −𝜋

4) = sin (

𝜋

4)



12. Prove that sin2 6𝑥 − sin2 4𝑥 = sin 2𝑥 sin 10𝑥

Solution:

Step1:

Given, L.H.S.= sin26𝑥 − sin24𝑥

= (sin6𝑥 − sin4𝑥)(sin6𝑥 + sin4𝑥)

sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2)and sin𝐴 − sin𝐵 = 2cos (

𝐴+𝐵

2) sin (

𝐴−𝐵

2)

L.H.S.= [2cos (6𝑥+4𝑥

2) sin (

6𝑥−4𝑥

2)] [2sin (

6𝑥+4𝑥

2) cos (

6𝑥−4𝑥

2)]

= [2 cos 5𝑥 sin 𝑥][2 sin 5𝑥 cos 𝑥]

= [2 sin 5𝑥 cos 5𝑥][2 sin 𝑥 cos 𝑥]

= sin 10𝑥 sin 2𝑥

=R.H.S.

Hence RHS = LHS

Overall Hint: Write the LHS using a^2 – b^2 = (a+b)*(a-b) then make use of the

formulae sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2)and sin𝐴 − sin𝐵 =

2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)

to prove the RHS

13. Prove that cos2 2𝑥 − cos2 6𝑥 = sin 4𝑥 sin 8𝑥

Solution:

Step1:

L.H.S.= cos22𝑥 − cos26𝑥

= (cos 2𝑥 + cos 6𝑥)(cos 2𝑥 − cos 6𝑥)

We know that,

cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵

2) sin (

𝐴−𝐵

2)andcos𝐴 + cos𝐵 = 2cos (

𝐴+𝐵

2) cos (

𝐴−𝐵

2)

So,L.H.S.= [−2sin (2𝑥+6𝑥

2) sin (

2𝑥−6𝑥

2)] [2cos (

2𝑥+6𝑥

2) cos (

2𝑥−6𝑥

2)]

= [−2sin(4𝑥)sin(−2𝑥)][2cos(4𝑥)cos(−2𝑥)]

= [2 sin 4𝑥 sin 2𝑥][2 cos 4𝑥 cos 2𝑥]

= [2 sin 4𝑥 cos 4𝑥][2 sin 2𝑥 cos 2𝑥]



= sin 8𝑥 sin 4𝑥

=R.H.S.

Hence LHS = RHS.

Overall Hint: Write the LHS in the form of a^2 – b^2 = (a+b)*(a-b) and make use of

the formulae cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵

2) sin (

𝐴−𝐵

2)andcos𝐴 + cos𝐵 =

2cos (𝐴+𝐵

2) cos (

𝐴−𝐵

2)

To prove the RHS

14. Prove that sin 2𝑥 + 2 sin 4𝑥 + sin 6𝑥 = 4 cos2 𝑥 sin 4𝑥

Solution:

Step1:

Given, L.H.S.= sin 2𝑥 + 2 sin 4𝑥 + sin 6𝑥

= (sin 2𝑥 + sin 6𝑥) + 2 sin 4𝑥

We know that sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2)

So,L.H.S.= 2sin (2𝑥+6𝑥

2) cos (

2𝑥−6𝑥

2) + 2 sin 4𝑥

Step2:

= 2 sin 4𝑥 cos(−2𝑥) + 2 sin 4𝑥

= 2 sin 4𝑥 [cos 2𝑥 + 1]

= 2 sin 4𝑥 [2cos2𝑥 − 1 + 1]

= 2 sin 4𝑥(2 cos2𝑥)

= 4cos2𝑥. sin 4𝑥

=R.H.S

Hence LHS = RHS.

Overall Hint: Write sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2) to prove the RHS

15. Prove that cot 4𝑥 (sin 5𝑥 + sin 3𝑥) = cot 𝑥 (sin 5𝑥 − sin 3𝑥)



Solution:

Step1:

Given, L.H.S.= cot 4𝑥 (sin 5𝑥 + sin 3𝑥)

We know that sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2)

So,L.H.S.=cos 4𝑥

sin 4𝑥[2sin (

5𝑥+3𝑥

2) cos (

5𝑥−3𝑥

2)]

=cos 4𝑥

sin 4𝑥(2 sin 4𝑥 cos 𝑥)

= 2 cos 4𝑥 cos 𝑥

R.H.S.= cot𝑥(sin5𝑥 − sin3𝑥)

Step2:

We know that sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)

So,R.H.S.=𝑐𝑜𝑠𝑥

sin𝑥[2cos (

5𝑥+3𝑥

2) sin (

5𝑥−3𝑥

2)]

=cos𝑥

sin𝑥[2cos4𝑥 sin𝑥]

= 2cos4𝑥 cos𝑥

L.H.S.=R.H.S.

Hence Proved

Overall Hint: Write cot 4𝑥 (sin 5𝑥 + sin 3𝑥) as cos 4𝑥

sin 4𝑥[2sin (

5𝑥+3𝑥

2) cos (

5𝑥−3𝑥

2)] using

formula sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2) in the LHS and cot 𝑥 (sin 5𝑥 −

sin 3𝑥) as 𝑐𝑜𝑠𝑥

sin𝑥[2cos (

5𝑥+3𝑥

2) sin (

5𝑥−3𝑥

2)] using the formula sin𝐴 − sin𝐵 =

2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)

In the RHS

16. Prove thatcos 9𝑥−cos 5𝑥

sin 17𝑥−sin 3𝑥= −

sin 2𝑥

cos 10𝑥

Solution:

Step1:



Given, L.H.S=cos 9𝑥−cos 5𝑥

sin 17𝑥−sin 3𝑥

We know that,

sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)andcos𝐴 − cos𝐵 = −2sin (

𝐴+𝐵

2) sin (

𝐴−𝐵

2)

L.H.S =− 2sin(

9𝑥+5𝑥

2)sin(

9𝑥−5𝑥

2)

2cos(17𝑥+3𝑥

2)sin(

17𝑥−3𝑥

2)

= −2sin 7𝑥. sin 2𝑥

2cos10𝑥. sin 7𝑥

= −sin 2𝑥

cos10𝑥

=R.H.S.

Hence LHS = RHS.

Overall Hint: Change the numerator and denominator in the LHS as cos 9𝑥−cos 5𝑥

sin 17𝑥−sin 3𝑥 =

− 2sin(9𝑥+5𝑥

2)sin(

9𝑥−5𝑥

2)

2cos(17𝑥+3𝑥

2)sin(

17𝑥−3𝑥

2) with the help of formula sin𝐴 − sin𝐵 =

2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)andcos𝐴 − cos𝐵 = −2sin (

𝐴+𝐵

2) sin (

𝐴−𝐵

2)

For the numerator and denominator respectively then simplify it to show that its equal to RHS

17. Prove that :sin5𝑥+sin3𝑥

cos5𝑥+cos3𝑥= tan4𝑥

Solution:

Step1:

L.H.S=sin5𝑥+sin3𝑥

cos5𝑥+cos3𝑥

We know that,


2) cos (

𝐴−𝐵


𝐴+𝐵

2) cos (

𝐴−𝐵

2)

Given, L.H.S = 2sin(

5𝑥+3𝑥

2)cos(

5𝑥−3𝑥

2)

2cos(5𝑥+3𝑥

2)cos(

5𝑥−3𝑥

2)

= 2sin(4𝑥)cos(𝑥)

2cos(4𝑥)cos(𝑥)

=sin4𝑥

cos4𝑥



= tan4𝑥

=R.H.S

Hence LHS = RHS

Overall Hint: Change the numerator and denominator of the LHS (sin5𝑥+sin3𝑥

cos5𝑥+cos3𝑥) as

2sin(5𝑥+3𝑥

2)cos(

5𝑥−3𝑥

2)

2cos(5𝑥+3𝑥

2)cos(

5𝑥−3𝑥

2) with the help of the formulae sin𝐴 + sin𝐵 =

2sin (𝐴+𝐵

2) cos (

𝐴−𝐵


𝐴+𝐵

2) cos (

𝐴−𝐵

2) then simplify it to

show that its equal to RHS

18. Prove that sin 𝑥−sin 𝑦

cos 𝑥+cos 𝑦= tan

x−y

2

Solution:

Step1:

Given, L.H.S=sin 𝑥−sin 𝑦

cos 𝑥+cos 𝑦

We know that sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2)andcos𝐴 + cos𝐵 =

2cos (𝐴+𝐵

2) cos (

𝐴−𝐵

2)

L.H.S = 2cos(

𝑥+𝑦

2)sin(

𝑥−𝑦

2)

2cos(𝑥+𝑦

2)cos(

𝑥−𝑦

2)

=sin (

𝑥−𝑦

2)

cos (𝑥−𝑦

2)

= tan (𝑥 − 𝑦

2)

=R.H.S.

Hence LHS = RHS

Overall Hint: Change the numerator and denominator of LHS (sin 𝑥−sin 𝑦

cos 𝑥+cos 𝑦) to

2cos(𝑥+𝑦

2)sin(

𝑥−𝑦

2)

2cos(𝑥+𝑦

2)cos(

𝑥−𝑦

2) using the formulae sin𝐴 − sin𝐵 = 2cos (

𝐴+𝐵

2) sin (

𝐴−𝐵

2)andcos𝐴 +

cos𝐵 = 2cos (𝐴+𝐵

2) cos (

𝐴−𝐵

2) and then simplify it to show that its equal to RHS



19. Prove thatsin𝑥+sin3𝑥

cos𝑥+cos3𝑥= tan 2𝑥

Solution:

Step 1:

Given, L.H.S.=sin𝑥+sin3𝑥

cos𝑥+cos3𝑥

=2sin(

𝑥+3𝑥

2)cos(

𝑥−3𝑥

2)

2cos(𝑥+3𝑥

2)cos(

𝑥−3𝑥

2) [∵ sin𝐴 + sin𝐵 = 2sin (

𝐴+𝐵

2) cos (

𝐴−𝐵

2)]

=sin 2𝑥

cos 2𝑥 [∵ cos𝐴 + cos𝐵 = 2cos (

𝐴+𝐵

2) cos (

𝐴−𝐵

2)]

= tan 2𝑥

=R.H.S.

Hence LHS = RHS.

Overall Hint: Change the numerator and denominator of the LHS ( sin𝑥+sin3𝑥

cos𝑥+cos3𝑥) to

2sin(𝑥+3𝑥

2)cos(

𝑥−3𝑥

2)

2cos(𝑥+3𝑥

2)cos(

𝑥−3𝑥

2) using the formulae [sin𝐴 + sin𝐵 = 2sin (

𝐴+𝐵

2) cos (

𝐴−𝐵

2)] and

[cos𝐴 + cos𝐵 = 2cos (𝐴+𝐵

2) cos (

𝐴−𝐵

2)] to prove the LHS equal to RHS

20. Prove that sin x−sin 3x

sin2 x−cos2 x= 2 sin x

Solution:

Step1:

Given, L.H.S.=sin𝑥−sin3𝑥

sin2𝑥−cos2𝑥

=2cos(

𝑥+3𝑥

2)sin(

𝑥−3𝑥

2)

− cos 2𝑥 [∵ sin𝐴 − sin𝐵 = 2cos (

𝐴+𝐵

2) sin (

𝐴−𝐵

2)]

=2 cos 2𝑥 sin(−𝑥)

− cos 2𝑥[∵ cos2𝑥 − sin2𝑥 = cos 2𝑥]

=−2 cos 2𝑥 sin 𝑥

− cos 2𝑥

= 2 sin 𝑥

=R.H.S.

Hence LHS = RHS.



Overall Hint: Change the numerator and denominator of LHS (sin x−sin 3x

sin2 x−cos2 x) to

2cos(𝑥+3𝑥

2)sin(

𝑥−3𝑥

2)

− cos 2𝑥 using the formulae [sin𝐴 − sin𝐵 = 2cos (

𝐴+𝐵

2) sin (

𝐴−𝐵

2)] and

[cos2𝑥 − sin2𝑥 = cos 2𝑥] 𝑎𝑛𝑑 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 𝑖𝑡 to prove the LHS equal to RHS

21. Prove thatcos4𝑥+cos3𝑥+cos2𝑥

sin4𝑥+sin3𝑥+sin2𝑥= cot 3𝑥

Solution:

Step1:

Given, L.H.S.=cos4𝑥+cos3𝑥+cos2𝑥

sin4𝑥+sin3𝑥+sin2𝑥

=(cos4𝑥 + cos2𝑥) + cos3𝑥

(sin4𝑥 + sin2𝑥) + sin3𝑥

We know that,


2) cos (

𝐴−𝐵


𝐴+𝐵

2) cos (

𝐴−𝐵

2)

=2cos (

4𝑥+2𝑥

2) cos (

4𝑥−2𝑥

2) + cos3𝑥

2sin (4𝑥+2𝑥

2) cos (

4𝑥−2𝑥

2) + sin3𝑥

Step2:

=2 cos 3𝑥 cos 𝑥 + cos 3𝑥

2 sin 3𝑥 cos 𝑥 + sin 3𝑥

=cos 3𝑥 (cos 𝑥 + 1)

sin 3𝑥 (cos 𝑥 + 1)

=cos 3𝑥

sin 3𝑥

= cot 3𝑥

=R.H.S.

Hence LHS = RHS

Overall Hint: Change the numerator and denominator of LHS( cos4𝑥+cos3𝑥+cos2𝑥

sin4𝑥+sin3𝑥+sin2𝑥) to

2cos(4𝑥+2𝑥

2)cos(

4𝑥−2𝑥

2)+cos3𝑥

2sin(4𝑥+2𝑥

2)cos(

4𝑥−2𝑥

2)+sin3𝑥

using the formulae sin𝐴 + sin𝐵 =

2sin (𝐴+𝐵

2) cos (

𝐴−𝐵


𝐴+𝐵

2) cos (

𝐴−𝐵

2) and simplify it to prove

the RHS



22. Prove that cot 𝑥 cot 2𝑥 − cot 2𝑥 cot 3𝑥 − cot 3𝑥 cot 𝑥 = 1

Solution:

Step1:

L.H.S.= cot 𝑥 cot 2𝑥 − cot 2𝑥 cot 3𝑥 − cot 3𝑥 cot 𝑥

= cot 𝑥 cot 2𝑥 − cot 3𝑥 (cot 2𝑥 + cot 𝑥)

= cot 𝑥 cot 2𝑥 − cot (2𝑥 + 𝑥)(cot2𝑥 + cot𝑥)

= cot 𝑥 cot 2𝑥 − [cot 2𝑥 cot 𝑥−1

cot 2𝑥+cot 𝑥] (cot 2𝑥 + cot 𝑥) [∵cot(𝐴 + 𝐵) =

[cot𝐴cot𝐵−1

cot𝐴+cot𝐵]

= cot 𝑥 cot 2𝑥 − cot 2𝑥 cot 𝑥 + 1

= 1

=R.H.S.

Hence LHS = RHS. Overall Hint: take cot3x in the 2nd and 3rd term and Use the formula cot(𝐴 + 𝐵) =

[cot𝐴cot𝐵−1

cot𝐴+cot𝐵] to show that LHS is equal to RHS

23. Prove that tan 4𝑥 =4 tan 𝑥(1−tan2 𝑥)

1−6 tan2 𝑥+tan4 𝑥

Solution:

Step1:

We know that,tan 2𝐴 =2 tan 𝐴

1−tan2𝐴

L.H.S.= tan4𝑥 = tan2(2𝑥)

=2 tan 2𝑥

1 − tan2(2𝑥)

=2 (

2 tan 𝑥

1−tan2𝑥)

1 − (2 tan 𝑥

1−tan2𝑥)

2

Step2:



=

4 tan 𝑥

1−tan2𝑥(1−tan2𝑥)2−4tan2𝑥

(1−tan2𝑥)2

=4 tan 𝑥(1−tan2𝑥)

(1−tan2𝑥)2−4tan2𝑥


1+tan4𝑥−2tan2𝑥−4tan2𝑥


1−6 tan2𝑥+tan4𝑥

=R.H.S.

Hence LHS = RHS

Overall Hint: Change the LHS tan4x as 2(

2 tan 𝑥

1−tan2𝑥)

1−(2 tan 𝑥

1−tan2𝑥)

2 using the formula tan 2A =

2 tan 𝐴

1−tan2𝐴 and then expand it to prove its equal to RHS

24. Prove thatcos 4𝑥 = 1 − 8 sin2 𝑥 cos2 𝑥

Solution:

Step1:

Given, L.H.S= cos 4𝑥 = cos2(2𝑥)

= 1 − 2sin2(2𝑥) [∵ cos2𝐴 = 1 − 2sin2𝐴]

= 1 − 2(2sin𝑥 cos𝑥)2] [∵ sin2𝐴 = 2sin𝐴cos𝐴

= 1 − 8sin2𝑥cos2𝑥

= R.H.S.

Hence LHS = RHS Overall Hint: Write cos(4x)= cos2(2x) and then apply formulae cos2𝐴 = 1 − 2sin2𝐴

sin2𝐴 = 2sin𝐴cos𝐴 to show it equal to RHS

25. Prove that cos 6𝑥 = 32 cos6 𝑥 − 48 cos4 𝑥 + 18 cos2 𝑥 − 1

Solution:



SteP1:

Given, L.H.S. = cos 6𝑥

= cos3(2𝑥)

= 4𝑐𝑜𝑠32𝑥 − 3 cos 2𝑥

[∵ cos 3𝐴 = 4cos3𝐴 − 3 cos 𝐴]

= 4[(2cos2𝑥 − 1)3 − 3(2cos2𝑥 − 1)]

[∵ 𝑐𝑜𝑠 2𝑥 = 2 cos2 𝑥 − 1]

Step2:

= 4[(2cos2𝑥)3 − (1)3 − 3(2cos2𝑥)2 + 3(2cos2𝑥)] − 6cos2𝑥 + 3

= 4[8cos6𝑥 − 1 − 12cos4𝑥 + 6cos2𝑥] − 6cos2𝑥 + 3

= 32 cos6𝑥 − 48 cos4𝑥 + 18 cos2𝑥 − 1

=R.H.S.

Hence LHS = RHS. Overall Hint: Change the terms in LHS by using formulae cos 3𝐴 = 4cos3𝐴 − 3 cos 𝐴

𝑐𝑜𝑠 2𝑥 = 2 cos2 𝑥 − 1 and simplify it to prove that its equal to RHS

Exercise: 3.4

Find the principal and general solutions of the following equations:

1. tan 𝑥 = √3

Solution:

Step1:

We know that tan𝜋

3= √3and tan

4𝜋

3= tan (𝜋 +

𝜋

3) = tan

𝜋

3= √3

Thus ,the principal solutions are 𝑥 = 𝜋

3and𝑥 =

4𝜋

3

And,tan𝑥 = tan𝜋

3

⇒ 𝑥 = 𝑛𝜋 +𝜋

3,where 𝑛 ∈ 𝑍

Thus, the general solution is= 𝑛𝜋 +𝜋

3, where 𝑛 ∈ 𝑍



Overall Hint : The principal solution is 𝜋 and 4𝜋

3 and the general solution is 𝑛𝜋 +

𝜋

3 , 𝑛 ∈ 𝑍

2. Find the principal and general solutions of the following equations:

sec 𝑥 = 2

Solution:

Step1:

Given, sec 𝑥 = 2

We know that sec𝜋

3= 2 andsec

5𝜋

3= sec (2𝜋 −

𝜋

3) = sec

𝜋

3= 2

Thus, the principal solutions are 𝑥 =𝜋

3and𝑥 =

5𝜋

3

Step2:

Now,sec 𝑥 = sec𝜋

3

⇒ cos 𝑥 = cos𝜋

3 [∵ sec 𝑥 =

1

cos 𝑥]

⇒ 𝑥 = 2𝑛𝜋 ±𝜋

3,where𝑛 ∈ 𝑍

Thus, the general solution is = 2𝑛𝜋 ±𝜋


Overall Hint: The principal solutions are 𝑥 =𝜋

3and𝑥 =

5𝜋

3 and the general solutions

are 2𝑛𝜋 ±𝜋

3, where 𝑛 ∈ 𝑍

3. Find the principal and general solutions of the following equations:

cot 𝑥 = −√3

Solution:

Step1:

We know that cot𝜋

6= √3

So,cot (𝜋 −𝜋

6) = −cot

𝜋

6= −√3andcot (2𝜋 −

𝜋

6) = −cot

𝜋

6= −√3

i.e.,cot (5𝜋

6) = −√3andcot (

11𝜋

6) = −√3



Thus, the principal solutions are 𝑥 =5𝜋

6and𝑥 =

11𝜋

6

Step2:

Now,cot 𝑥 = cot (5𝜋

6)

⇒ tan 𝑥 = tan5𝜋

6[tan 𝑥 =

1

cot 𝑥]

⇒ 𝑥 = 𝑛𝜋 +5𝜋


Thus, the general solution is= 𝑛𝜋 +5𝜋


Overall Hint: The principal solution is 𝑥 =5𝜋

6and𝑥 =

11𝜋

6 and the general solution is

𝑛𝜋 +5𝜋


4. Find the principal and general solution

cosec 𝑥 = −2

Solution:

Step1:

cosec 𝑥 = −2

We know that cosec𝜋

6= 2

Also, cosec (𝜋 +𝜋

6) = −cosec

𝜋

6= −2andcosec (2𝜋 −

𝜋

6) = −cosec

𝜋

6= −2

i.e., cosec (7𝜋

6) = −2andcosec (

11𝜋

6) = −2

Thus, the principal solutions are 𝑥 =7𝜋

6and𝑥 =

11𝜋

6

Step2:

Now,cosec 𝑥 = cosec (7𝜋

6)

⇒ sin𝑥 = sin7𝜋

6[sin 𝑥 =

1

cosec 𝑥]

⇒ 𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋


Thus, the general solution is= 𝑛𝜋 + (−1)𝑛 7𝜋

6 , where 𝑛 ∈ 𝑍



Overall Hint: The principal solution is 𝑥 =7𝜋

6and𝑥 =

11𝜋

6 and the general solution is

𝑛𝜋 + (−1)𝑛 7𝜋


5. Find the general solution for each of the following equations:

cos 4𝑥 = cos 2𝑥

Solution:

Step1:

Given, cos4𝑥 = cos2𝑥

⇒ cos4𝑥 − cos2𝑥 = 0

cos4𝑥 − cos2𝑥 = −2sin (4𝑥+2𝑥

2) sin (

4𝑥−2𝑥

2)

[∵ cos𝐴 − cos𝐵 = −2sin (𝐴 + 𝐵

2) sin (

𝐴 − 𝐵

2)]

Step2:

⇒ −2 sin 3𝑥 sin 𝑥 = 0

⇒ sin3𝑥 = 0 orsin𝑥 = 0

⇒ 3𝑥 = 𝑛𝜋 or 𝑥 = 𝑛𝜋, where n ∈ Z

Thus, 𝑥 =𝑛𝜋

3or 𝑥 = 𝑛𝜋, where n ∈ Z

Overall Hint: The general solution is 𝑥 =𝑛𝜋

3or 𝑥 = 𝑛𝜋, where n ∈ Z

Use cos𝐴 − cos𝐵 = −2sin (𝐴+𝐵

2) sin (

𝐴−𝐵

2)


cos 3𝑥 + cos 𝑥 − cos 2𝑥 = 0

Solution:

Step1:

Given, cos 3𝑥 + cos 𝑥 − cos 2𝑥 = 0



⇒ 2cos (3𝑥 + 𝑥

2) cos (

3𝑥 − 𝑥

2) − cos 2𝑥 = 0

[∵ cos𝐴 + cos𝐵 = 2cos (𝐴 + 𝐵

2) cos (

𝐴 − 𝐵

2)]

Step2:

⇒ 2 cos 2𝑥 cos 𝑥 − cos 2𝑥 = 0

⇒ cos 2𝑥 (2 cos 𝑥 − 1) = 0

⇒ cos 2𝑥 = 0 or 2 cos 𝑥 − 1 = 0

⇒ cos 2𝑥 = 0 or cos 𝑥 =1

2

Therefore, 2𝑥 = (2𝑛 + 1)𝜋

2orcos 𝑥 = cos

𝜋

3, where n ∈ Z

Thus,𝑥 = (2𝑛 + 1)𝜋

4or 𝑥 = 2𝑛𝜋 ±

𝜋

3,where n ∈ Z

Overall Hint: The general solution is 𝑥 = (2𝑛 + 1)𝜋

4or 𝑥 = 2𝑛𝜋 ±

𝜋

3,where n ∈ Z

Use cos𝐴 + cos𝐵 = 2cos (𝐴+𝐵

2) cos (

𝐴−𝐵

2)


sin 2𝑥 + cos 𝑥 = 0

Solution:

Step1:

Given, sin 2𝑥 + cos 𝑥 = 0

⇒ 2 sin 𝑥 cos 𝑥 + cos 𝑥 = 0

[∵ sin𝐴𝑥 = 2sin𝐴 cos𝐴]

⇒ cos 𝑥 (2 sin 𝑥 + 1) = 0

⇒ cos 𝑥 = 0 or 2 sin 𝑥 + 1 = 0

Step2:

⇒ cos 𝑥 = 0 ⇒ 𝑥 = (2𝑛 + 1)𝜋

2,where n ∈ Z

Also,2sin𝑥 + 1 = 0

⇒ sin𝑥 = −1

2



⇒ sin𝑥 = −sin𝜋

6 ⇒ sin 𝑥 = sin (𝜋 +

𝜋

6) = sin

7𝜋

6

⇒ 𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋

6,where n ∈ Z

Thus, the general solution is 𝑥 = (2𝑛 + 1)𝜋

2or𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋

6,where n ∈ Z

Overall Hint: The general solution is 𝑥 = (2𝑛 + 1)𝜋

2or𝑥 = 𝑛𝜋 + (−1)𝑛 7𝜋

6,where n ∈

Z ,Use sin𝐴𝑥 = 2sin𝐴 cos𝐴

8. Find the general solution for each of the following equations

sec2 2𝑥 = 1 − tan 2𝑥

Solution:

Step1:

Given, sec22𝑥 = 1 − tan 2𝑥

1 + tan22𝑥 = 1 − tan 2𝑥

⇒ tan22𝑥 = − tan 2𝑥

⇒ tan22𝑥 + tan 2𝑥 = 0

⇒ tan 2𝑥 (tan 2𝑥 + 1) = 0

⇒ tan 2𝑥 = 0 Or tan 2𝑥 + 1 = 0

Now, tan 2𝑥 = tan 0

⇒ 2𝑥 = 𝑛𝜋 + 0, Where n ∈ Z

Step2:

⇒𝑥 =𝑛𝜋

2, where n ∈ Z

tan 2𝑥 + 1 = 0

⇒ tan 2𝑥 = −1

⇒ tan 2𝑥 = −tan𝜋

4 ⇒ tan 2𝑥 = tan (𝜋 −

𝜋

4) = tan

3𝜋

4

So,2𝑥 = 𝑛𝜋 +3𝜋

4, where n ∈ Z

⇒ 𝑥 =𝑛𝜋

2+

3𝜋

8, where n ∈ Z



Thus, the general solution is 𝑥 =𝑛𝜋

2or 𝑥 =

𝑛𝜋

2+

3𝜋

8, where n ∈ Z

Overall Hint: The general solution is 𝑥 =𝑛𝜋

2or 𝑥 =

𝑛𝜋

2+

3𝜋

8, where n ∈ Z , Use sec22𝑥

=

1 + tan22𝑥. Take tan2x common.

9. Find the general solution for each of the following equation

sin 𝑥 + sin 3𝑥 + sin 5𝑥 = 0

Solution:

Step1:

Given, sin 𝑥 + sin 3𝑥 + sin 5𝑥 = 0

(sin𝑥 + sin5𝑥) + sin3𝑥 = 0

⇒ 2sin (𝑥+5𝑥

2) cos (

𝑥−5𝑥

2) + sin3𝑥 = 0

[∵ sin𝐴 + sin𝐵 = 2sin (𝐴 + 𝐵

2) cos (

𝐴 − 𝐵

2)]

⇒ 2 sin 3𝑥 cos(−2𝑥) + sin 3𝑥 = 0

⇒ sin 3𝑥 (2 cos 2𝑥 + 1) = 0

⇒ sin 3𝑥 = 0 or2 cos 2𝑥 + 1 = 0

⇒ sin3𝑥 = 0 or cos 2𝑥 = −1

2

sin 3𝑥 = 0

Step2:

3𝑥 = 𝑛𝜋, where n ∈ Z

⇒ 𝑥 =𝑛𝜋

3, where n ∈ Z

cos 2𝑥 = −1

2

cos 2𝑥 = −cos𝜋

3⇒ cos 2𝑥 = cos (𝜋 −

𝜋

3) = cos

2𝜋

3

So,2𝑥 = 2𝑛𝜋 ±2𝜋

3, where n ∈ Z



⇒ 𝑥 = 𝑛𝜋 ±𝜋

3, where n ∈ Z

Thus, the general solution is𝑥 =𝑛𝜋

3or 𝑥 = 𝑛𝜋 ±

𝜋

3, where n ∈ Z

Overall Hint: The general solution is𝑥 =𝑛𝜋

3or 𝑥 = 𝑛𝜋 ±

𝜋

3, where n ∈ Z , Use sin𝐴 +

sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2)

Miscellaneous Exercise

1. Prove that:2 cosπ

13cos

9π

13+ cos

3π

13+ cos

5π

13= 0

Solution:

Step1:

Given, L.H.S.= 2cos𝜋

13cos

9𝜋

13+ cos

3𝜋

13+ cos

5𝜋

13

2cos𝜋

13cos

9𝜋

13+ 2cos (

3𝜋

13+

5𝜋

13

2) cos (

3𝜋

13−

5𝜋

13

2)

[cos 𝑥 + cos 𝑦 = 2 cos (𝑥 + 𝑦

2) cos (

𝑥 − 𝑦

2)]

⇒ 2cos𝜋

13cos

9𝜋

13+ 2cos (

4𝜋

13) cos (−

𝜋

13)

Step2:

⇒ 2cos𝜋

13cos

9𝜋

13+ 2cos (

4𝜋

13) cos (

𝜋

13)

⇒ 2cos𝜋

13(cos

9𝜋

13+ cos (

4𝜋

13))

= 2cos𝜋

13(2cos (

9𝜋

13+

4𝜋

13

2) cos (

9𝜋

13−

4𝜋

13

2))

= 4cos𝜋

13(cos (

𝜋

2) cos (

5𝜋

26))

= 4cos𝜋

13(0 × cos (

5𝜋

26))



= 0

=R.H.S.

Hence LHS = RHS.

Overall Hint: Use cos 𝑥 + cos 𝑦 = 2 cos (𝑥+𝑦

2) cos (

𝑥−𝑦

2) in the :LHS to prove it is

equal to RHS

2. Prove that:(sin 3𝑥 + sin 𝑥) sin 𝑥 + (cos 3𝑥 – cos 𝑥) cos 𝑥 = 0

Solution:

Step1:

Given, L.H.S.= (sin 3𝑥 + sin 𝑥) sin 𝑥 + (cos 3𝑥 − cos 𝑥) cos 𝑥

= sin 3𝑥 sin 𝑥 + sin2𝑥 + cos 3𝑥 cos 𝑥 − cos2 𝑥

= (sin 3𝑥 sin 𝑥 + cos 3𝑥 cos 𝑥)+(sin2𝑥 − cos2𝑥)

= cos(3𝑥 − 𝑥) − cos 2𝑥) [∵ cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵 &cos2𝐴 =cos2𝐴 − sin2𝐴 ]

= cos2𝑥 − cos2𝑥

= 0

=R.H.S.

Hence LHS = RHS..

Overall Hint: Use cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵 &cos2𝐴 = cos2𝐴 − sin2𝐴 to prove LHS is equal to RHS

3. Prove that:(cos 𝑥 + cos 𝑦)2 + (sin 𝑥 − sin 𝑦)2 = 4 cos2 x+𝑦

2

Solution:

Step1:

Given, L.H.S.= (cos𝑥 + cos𝑦)2 + (sin𝑥 − sin𝑦)2

= cos2𝑥 + cos2𝑦 + 2 cos 𝑥 cos 𝑦 + sin2𝑥 + sin2𝑦 − 2 sin 𝑥 sin 𝑦

= (sin2𝑥 + cos2𝑥)+(sin2𝑦 + cos2𝑦) + 2(cos 𝑥 cos 𝑦 − sin 𝑥 sin 𝑦)



= (1) + (1) + 2cos (𝑥 + 𝑦) [∵ cos(𝐴 + 𝐵) = cos𝐴 cos𝐵 − sin𝐴 sin𝐵]

Step2:

= 2[1 + cos(𝑥 + 𝑦)]

= 2 [2cos2 (𝑥+𝑦

2)] [ ∵ 1 + cos 2𝐴 = 2cos2𝐴 ]

= 4cos2 (𝑥 + 𝑦

2)

=R.H.S.

Hence LHS = RHS. Overall Hint: Use cos(𝐴 + 𝐵) = cos𝐴 cos𝐵 − sin𝐴 sin𝐵 , 1 + cos 2𝐴 = 2cos2𝐴 , (a+b)^2 = (a^2+b^2+2ab) , (a-b)^2 = (a^2+b^2-2ab) to prove LHS is equal to RHS

4. Prove that:(cos 𝑥 − cos 𝑦)2 + (sin 𝑥 − sin 𝑦)2 = 4 sin2 𝑥−𝑦

2

Solution:

Step1:

Given, L.H.S.= (cos 𝑥 − cos 𝑦)2 + (sin 𝑥 − sin 𝑦)2

= cos2𝑥 + cos2𝑦 − 2 cos 𝑥 cos 𝑦 + sin2𝑥 + sin2𝑦 − 2 sin 𝑥 sin 𝑦

= (sin2𝑥 + cos2𝑥)+(sin2𝑦 + cos2𝑦) − 2(cos𝑥 cos𝑦 + sin𝑥 sin𝑦)

= (1) + (1) − 2 cos(𝑥 − 𝑦)

Step2:

[∵ cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵]

= 2[1 − cos(𝑥 − 𝑦)]

= 2 [1 − (1 − 2sin2 (𝑥−𝑦

2))] [∵ 1 − cos2𝐴 = 2sin2𝐴]

= 4sin2 (𝑥 − 𝑦

2)

=R.H.S.

Hence LHS = RHS. Overall Hint: Use (a-b)^2 = (a^2+b^2-2ab) , 1 − cos2𝐴 = 2sin2𝐴 ,



cos(𝐴 − 𝐵) = cos𝐴 cos𝐵 + sin𝐴 sin𝐵 to prove LHS equal to RHS

5. Prove that :sin 𝑥 + sin 3𝑥 + sin 5𝑥 + sin 7𝑥 = 4 cos 𝑥 cos 2𝑥 sin 4𝑥

Solution:

Step1:

Given, L.H.S.= sin 𝑥 + sin 3𝑥 + sin 5𝑥 + sin 7𝑥

⇒ (sin 𝑥 + sin 5𝑥) + (sin 3𝑥 + sin 7𝑥)

⇒ 2sin (𝑥 + 5𝑥

2) cos (

𝑥 − 5𝑥

2) + 2sin (

3𝑥 + 7𝑥

2) cos (

3𝑥 − 7𝑥

2)


2) cos (

𝐴 − 𝐵

2)]

⇒ 2 sin 3𝑥 cos(−2𝑥) + 2 sin 5𝑥 cos(−2𝑥)

Step2:

= 2 sin 3𝑥 cos2𝑥 + 2 sin 5𝑥 cos 2𝑥

= 2 cos 2𝑥 (sin 3𝑥 + sin 5𝑥)

= 2 cos 2𝑥 [2sin (3𝑥 + 5𝑥

2) cos (

3 𝑥 − 5𝑥

2)]

= 4 cos 2𝑥 sin 4𝑥 cos(−𝑥)

= 4 cos 2𝑥 sin 4𝑥 cos 𝑥

=R.H.S.

Hence RHS = LHS

Overall Hint: Use sin𝐴 + sin𝐵 = 2sin (𝐴+𝐵

2) cos (

𝐴−𝐵

2) formula to show that the

LHS is equal to RHS

6. Prove that (sin7𝑥+sin5𝑥)+(sin9𝑥+sin3𝑥)

(cos7𝑥+cos5𝑥)+(cos9𝑥+cos3𝑥)= tan6x

Solution:

Given, L.H.S.=(sin7𝑥+sin5𝑥)+(sin9𝑥+sin3𝑥)

(cos7𝑥+cos5𝑥)+(cos9𝑥+cos3𝑥)

We know that,



sin 𝐴 + sin 𝐵 = 2sin (𝐴 + 𝐵

2) cos (

𝐴 − 𝐵

2) & cos 𝐴 + cos 𝐵

= 2cos (𝐴 + 𝐵

2) cos (

𝐴 − 𝐵

2)

L.H.S. ={2sin(

7𝑥+5𝑥

2)cos(

7𝑥−5𝑥

2)}+{2sin(

9𝑥+3𝑥

2)cos(

9𝑥−3𝑥

2)}

{2cos(7𝑥+5𝑥

2)cos(

7𝑥−5𝑥

2)}+{2cos(

9𝑥+3𝑥

2)cos(

9𝑥−3𝑥

2)}

Step2:

=2 sin 6𝑥 cos 𝑥 + 2 sin 6𝑥 cos 3𝑥

2 cos 𝑥 cos 𝑥 + 2 cos 6𝑥 cos 3𝑥

=2 sin 6𝑥 (cos 𝑥 + cos 3𝑥)

2 cos 6𝑥 (cos 𝑥 + cos 3𝑥)

=sin 6𝑥

cos 6𝑥

= tan 6𝑥

=R.H.S.

Hence LHS = RHS.

Overall Hint: Change the LHS( (sin7𝑥+sin5𝑥)+(sin9𝑥+sin3𝑥)

(cos7𝑥+cos5𝑥)+(cos9𝑥+cos3𝑥) to

{2sin(7𝑥+5𝑥

2)cos(

7𝑥−5𝑥

2)}+{2sin(

9𝑥+3𝑥

2)cos(

9𝑥−3𝑥

2)}

{2cos(7𝑥+5𝑥

2)cos(

7𝑥−5𝑥

2)}+{2cos(

9𝑥+3𝑥

2)cos(

9𝑥−3𝑥

2)}

using formula

sin 𝐴 + sin 𝐵 = 2sin (𝐴 + 𝐵

2) cos (

𝐴 − 𝐵

2) & cos 𝐴 + cos 𝐵

= 2cos (𝐴 + 𝐵

2) cos (

𝐴 − 𝐵

2)

And show it is equal to RHS

7. Prove that: sin 3𝑥 + sin 2𝑥 − sin 𝑥 = 4 sin 𝑥 cos𝑥

2cos

3𝑥

2

Solution:

Step1:

Given, L.H.S.sin 3𝑥 + sin 2𝑥 − sin 𝑥 = sin 3𝑥 + sin 2𝑥 − sin 𝑥

= sin 3𝑥 + 2cos (2𝑥 + 𝑥

2) sin (

2𝑥 − 𝑥

2)

[ ∵ sin𝐴 − sin𝐵 = 2cos (𝐴 + 𝐵

2) sin (

𝐴 − 𝐵

2)]



= sin 3𝑥 + 2cos3𝑥

2sin

𝑥

2

= 2sin3𝑥

2cos

3𝑥

2+ 2cos

3𝑥

2sin

𝑥

2

= 2cos3𝑥

2(sin

3𝑥

2+ sin

𝑥

2)

= 2cos3𝑥

2(2sin (

3𝑥

2+

𝑥

2

2) cos (

3𝑥

2−

𝑥

2

2))

Step2:


2) cos (

𝐴 − 𝐵

2)]

= 2cos3𝑥

2(2sin𝑥 cos (

𝑥

2))

= 4 sin 𝑥 cos𝑥

2cos

3𝑥

2

=R.H.S.

Hence LHS = RHS.

Overall Hint: Change LHS sin 3𝑥 + sin 2𝑥 − sin 𝑥 to sin 3𝑥 + 2cos (2𝑥+𝑥

2) sin (

2𝑥−𝑥

2)

by using the formula

Sin𝐴 − sin𝐵 = 2cos (𝐴+𝐵

2) sin (

𝐴−𝐵

2) use sin𝐴 + sin𝐵 = 2sin (

𝐴+𝐵

2) cos (

𝐴−𝐵

2) for

similarly changing the RHS and showing they both are equal

8. Find sin𝑥

2, cos

𝑥

2andtan

𝑥

2intan 𝑥 = −

4

3, 𝑥 in quadrant II

Solution:

Step1:

Here, 𝑥 lies in second quadrant.

So,𝜋

2< 𝑥 < 𝜋

⇒𝜋

4<

𝑥

2<

𝜋

2



Therefore, sin𝑥

2, cos

𝑥

2, tan

𝑥

2liesinfirstquadrant so they all are positive.

Given,tan 𝑥 = −4

3

We know that ,tan 2𝐴 =2tan𝐴

1−tan2𝐴

So,tan𝑥 = 2tan

𝑥

2

1−tan2𝑥

2

⇒ −4

3=

2tan𝑥

2

1 − tan2 𝑥

2

⇒ −4 (1 − tan2𝑥

2) = 6tan

𝑥

2

Step2:

⇒ 4tan2𝑥

2− 6tan

𝑥

2− 4 = 0

⇒ 2tan2𝑥

2− 3tan

𝑥

2− 2 = 0

⇒ 2tan2𝑥

2− 4tan

𝑥

2+ tan

𝑥

2− 2 = 0

⇒ 2tan𝑥

2(tan

𝑥

2− 2) + 1 (tan

𝑥

2− 2) = 0

⇒ (tan𝑥

2− 2) (2tan

𝑥

2+ 1) = 0

tan𝑥

2= 2ortan

𝑥

2= −

1

2

tan𝑥

2= 2 [tan

𝑥

2 Cannot be negative as

𝑥

2 is in first quadrant]

Step3:

sec2𝑥

2= 1 + tan2

𝑥

2

sec2𝑥

2= 1 + (2)2

⇒ sec2𝑥

2= 5

⇒ sec𝑥

2= √5 [sec

𝑥

2 Cannot be negative as

𝑥

2 is in first quadrant]

⇒ cos𝑥

2=

1

√5



Therefore, tan𝑥

2= 2

sin𝑥

2

cos𝑥

2

= 2

⇒ sin𝑥

2= 2cos

𝑥

2

Therefore, sin𝑥

2=

2

√5

Overall Hint: Find tan𝑥

2 using tan2x =

2tan𝑥

2

1−tan2𝑥

2

then find similarly values of

sin𝑥

2, cos

𝑥

2 using sin

𝑥

2= tan

𝑥

2cos

𝑥

2 and cos

𝑥

2 = sin

𝑥

2/ tan

𝑥

2

9. Find sin𝑥

2, cos

𝑥

2 and 𝑡𝑎𝑛

𝑥

2 for cos 𝑥 = −

1

3, 𝑥 in quadrant III

Solution:

Step1:

Here,𝑥liesinthirdquadrant.

So,𝜋 < 𝑥 <3𝜋

2

⇒𝜋

2<

𝑥

2<

3𝜋

4

It is known that sin𝑥

2is positive whereas, cos

𝑥

2andtan

𝑥

2are negative, lies in third

quadrant.

Given,cos 𝑥 = −1

3

We know that,cos 2𝐴 = 2cos2𝐴 − 1

So,cos 𝑥 = 2cos2 𝑥

2− 1

⇒ −1

3= 2cos2

𝑥

2− 1

⇒ cos2𝑥

2=

1 + cos 𝑥

2=

1 + (−1

3)

2=

(2

3)

2=

1

3

⇒ cos𝑥

2= ±

1

√3



Step2:

Whereas, cos𝑥

2 is negative as

𝑥

2 lies in second quadrant.

So,cos𝑥

2= −

1

√3

We know that, sin2𝐴 + cos2𝐴 = 1

sin2𝑥

2+ (

1

√3)

2

= 1

sin2𝑥

2= 1 −

1

3

⇒ sin2𝑥

2=

2

3

sin𝑥

2= ±√

2

3

Step3:

But sin𝑥

2 is positive as

𝑥

2 lies in second quadrant.

So,sin𝑥

2= √

2

3

Also,tan𝑥

2=

sin𝑥

2

cos𝑥

2

tan𝑥

2=

√2

3

−1

√3

Therefore, tan𝑥

2= −√2

Overall Hint: Find

cos𝑥

2 using

𝑐𝑜𝑠 2𝐴= 2cos2𝐴 − 1 and then the values of

sin𝑥

2= √1– cos2 𝑥

2

𝑡𝑎𝑛𝑥

2 using tan𝑥 =

2tan𝑥

2

1−tan2𝑥

2



10. Find sin𝑥

2, cos

𝑥

2 and tan

𝑥

2 for sin 𝑥 =

1

4, 𝑥 in quadrant II.

Solution:

Step1:

Here,𝑥 lies in second quadrant.

So,𝜋

2< 𝑥 < 𝜋

⇒𝜋

4<

𝑥

2<

𝜋

2

That means sin𝑥

2, cos

𝑥

2, tan

𝑥

2 lies in first quadrant.

Given, sin 𝑥 =1

4

We know that, sin2𝐴 + cos2𝐴 = 1

sin2𝑥 + cos2𝑥 = 1

= cos2𝑥 = 1 − (1

4)

2

⇒ cos2𝑥 = 1 −1

16=

15

16

cos 𝑥 = ±√15

4

Step2:

But cos 𝑥 is negative as 𝑥 lies in second quadrant.

So,cos𝑥 = −√15

4

sin2𝑥

2=

1 − cos𝑥

2[∵ cos2𝐴 = 1 − 2sin2𝐴]

sin2𝑥

2=

1 +√15

4

2

sin2𝑥

2=

4 + √15

8

sin𝑥

2= √(

4+√15

8) [sin

𝑥

2is positive as

𝑥

2lies in first quadrant]

cos2 𝑥

2=

1+cos𝑥

2 [∵ cos2𝐴 = 1 − 2sin2𝐴]



cos2𝑥

2=

1 −√15

4

2

cos2𝑥

2=

4 − √15

8

cos𝑥

2= √(

4−√15

8) [cos

𝑥

2 Is positive as

𝑥

2 lies in first quadrant]

Step3:

Also,tan𝑥

2=

sin𝑥

2

cos𝑥

2

tan𝑥

2=

√(4+√15

8)

√(4−√15

8)

⇒ tan𝑥

2=

√4 + √15

√4 − √15

⇒ √4 + √15

4 − √15×

4 + √15

4 + √15

⇒ √(4 + √15)2

16 − 15

⇒ 4 + √15

Therefore, the values of sin𝑥

2 , 𝑐𝑜𝑠

𝑥

2, tan

𝑥

2 are √(

4+√15

8), √(

4−√15

8) and 4 + √15.

Overall Hint: First find cos 𝑥 using the formula sin2𝐴 + cos2𝐴 = 1 then proceed accordingly

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Class XI CBSE-Mathematics Trigonometric Functions...Class–XI–CBSE-Mathematics Trigonometric...

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