Prove that: sin 2 72 ° - cos 2 30 ° = 5 - 1 8 .

To prove sin 2 72 ° - cos 2 30 ° = 5 - 1 8 LHS = sin 2 72 ° - cos 2 30 ° = cos 2 18 ° - 3 4 ∵ sin 72 ° = sin 90 ° - 18 ° = cos 18 ° = 1 - sin 2 18 ° - 3 4 ∵ sin 2 x + cos 2 x = 1 = 1 4 - 5 - 1 4 2 = 4 16 - ( 5 + 1 - 2 5 ) 16 = - 2 + 2 5 16 = 5 - 1 8 = RHS Hence, proved.

Prove that: sin 6 ° sin 42 ° sin 66 ° sin 78 ° = 1 16 .

To prove, sin 6 ° sin 42 ° sin 66 ° sin 78 ° = 1 16 LHS = sin 6 ° sin 42 ° sin 66 ° sin 78 ° = sin 6 ° sin 90 ° - 48 ° sin 90 ° - 24 ° sin 90 ° - 12 ° = sin 6 ° cos 48 ° cos 24 ° cos 12 ° ∵ sin 90 ° - θ = cos θ = 1 2 cos 6 ° 2 sin 6 ° cos 6 ° cos 12 ° cos 24 ° cos 48 ° = 1 4 cos 6 ° 2 sin 12 ° cos 12 ° cos 24 ° cos 48 ° ∵ 2 sin x cos x = sin 2 x = 1 4 cos 6 ° 2 sin 24 ° cos 24 ° cos 48 ° = 1 16 cos 6 ° 2 sin 48 ° cos 48 ° = sin 96 ° 16 cos 6 ° = sin 90 ° + 6 ° 16 cos 6 ° = cos 6 ° 16 cos 6 ° ∵ sin ( 90 ° + θ ) = cos θ = 1 16 = RHS Hence, proved.

Prove that: cos π 5 cos 2 π 5 cos 4 π 5 cos 8 π 5 = - 1 16 .

To prove, cos π 5 cos 2 π 5 cos 4 π 5 cos 8 π 5 = - 1 16 LHS = cos π 5 cos 2 π 5 cos 4 π 5 cos 8 π 5 = 1 2 sin π 5 2 sin π 5 cos π 5 cos 2 π 5 cos 4 π 5 cos 8 π 5 = 1 4 sin π 5 2 sin 2 π 5 cos 2 π 5 cos 4 π 5 cos 8 π 5 ∵ 2 sin x cos x = sin 2 x = 1 8 sin π 5 2 sin 4 π 5 cos 4 π 5 cos 8 π 5 = 1 16 sin π 5 2 sin 8 π 5 cos 8 π 5 = sin 16 π 5 16 sin π 5 = sin 3 π + π 5 16 sin π 5 = - sin π 5 16 sin π 5 ∵ sin ( n π + θ ) = ( - 1 ) n sin θ , n ∈ Z = - 1 16 = RHS Hence, proved.

Find sin x 2 , cos x 2 and tan x 2 if tan x = - 4 3 , x lies in second quadrant.

Given that, x lies in II quadrant ⇒ x ∈ π 2 , π ⇒ x 2 ∈ π 4 , π 2 ⇒ x 2 lies in I quadrant. All trigonometric ratios of x 2 are positive. Given that, tan x = - 4 3 ⇒ 2 tan x 2 1 - tan 2 x 2 = - 4 3 ⇒ 6 tan x 2 = - 4 + 4 tan 2 x 2 ⇒ 2 2 tan 2 x 2 - 3 tan x 2 - 2 = 0 ⇒ 2 tan 2 x 2 - 4 tan x 2 + tan x 2 - 2 = 0 ⇒ 2 tan x 2 tan x 2 - 2 + 1 tan x 2 - 2 = 0 ⇒ 2 tan x 2 + 1 tan x 2 - 2 = 0 ⇒ tan x 2 = 2 ∵ tan x 2 ≠ - 1 2 , a s x 2 ∈ I quadrant ⇒ sin x 2 cos x 2 = 2 ⇒ sin x 2 = 2 cos x 2 ...(i) As tan x = - 4 3 ⇒ sin x cos x = - 4 3 ⇒ 2 sin x 2 cos x 2 2 cos 2 x 2 - 1 = - 4 3 [from (i)] ⇒ 4 cos 2 x 2 2 cos 2 x 2 - 1 = - 4 3 ⇒ 3 cos 2 x 2 = - 2 cos 2 x 2 + 1 ⇒ cos 2 x 2 = 1 5 ⇒ cos x 2 = 1 5 is positive, since x 2 ∈ I quadrant. Now, sin x 2 = 2 cos x 2 ⇒ sin x 2 = 2 5 ∴ sin x 2 = 2 5 , cos x 2 = 1 5 and tan x 2 = 2 .

Find sin x 2 , cos x 2 and tan x 2 if cos x = - 1 3 , x lies in third quadrant.

Given that, x lies in III quadrant. ⇒ x ∈ π , 3 π 2 ⇒ x 2 ∈ π 2 , 3 π 4 ⇒ x 2 lies in second quadrant. ⇒ Only sin x 2 and cosec x 2 are positive. Also given that, cos x = - 1 3 ⇒ 2 cos 2 x 2 - 1 = - 1 3 ⇒ cos 2 x 2 = 1 2 1 - 1 3 ⇒ cos 2 x 2 = 1 3 ⇒ cos x 2 = - 1 3 [ ∵ x 2 ∈ II quadrant ] Again, cos x = - 1 3 ⇒ 1 - 2 sin 2 x 2 = - 1 3 ⇒ 1 + 1 3 1 2 = sin 2 x 2 ⇒ sin 2 x 2 = 2 3 ⇒ sin x 2 = 2 3 Now, tan x 2 = sin x 2 cos x 2 ⇒ tan x 2 = 2 3 × - 3 = - 2 ∴ sin x 2 = 2 3 , cos x 2 = - 1 3 and tan x 2 = - 2 .

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7

M L Aggarwal Solutions for Chapter: Trigonometry, Exercise 7: EXERCISE 3.7

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Trigonometry, Exercise 7: EXERCISE 3.7

Attempt the practice questions on Chapter 3: Trigonometry, Exercise 7: EXERCISE 3.7 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Trigonometry, Exercise 7: EXERCISE 3.7 with Hints & Solutions

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7>Q 21

Prove that: $\sin^{2} 72 ° - \cos^{2} 30 ° = \frac{\sqrt{5} - 1}{8}$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7>Q 22

Prove that: $\sin 6 ° \sin 42 ° \sin 66 ° \sin 78 ° = \frac{1}{16}$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7>Q 22

Prove that: $\cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} = - \frac{1}{16}$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7>Q 23

Given that $\cos \frac{x}{2} = \frac{12}{13}$ and $x$ lies in first quadrant, calculate without the use of tables, the values of $\sin x, \cos x$ and $\tan x$ .

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7>Q 24

Given that $\tan x = \frac{12}{5}, \cos y = - \frac{3}{5}$ and $x, y$ are in the same quadrant, calculate without the use of tables the value of $\sin (x + y)$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 3 - Trigonometry>EXERCISE 3.7>Q 24

Given that $\tan x = \frac{12}{5}, \cos y = - \frac{3}{5}$ and $x, y$ are in the same quadrant, calculate without the use of tables the value of $\cos \frac{y}{2}$ .