AP Inter 1st Year Maths Exercise 3c Solutions

Referring to the AP Inter 1st Year Maths Study Material Chapter Trigonometric Functions Exercise 3c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Trigonometric Functions Exercise 3c

I.
Question 1.
Find the value of :

i) sin 75°
Solution:
i) sin 75° = sin (45 + 30)°
= sin 45°°cos 30° + cos°45°sin 30°
[∵ sin (A + B) = sin A cos B + cos A sin B]
= \(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

ii) tan 15°
Solution:
tan 15° = tan (45° – 30°)

iii) cot 15°
Solution:
cot 15° = \(\frac{1}{\tan 15^{\circ}}\)
= \(\frac{1}{2-\sqrt{3}}\)
= \(\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}\)
= 2 + √3

iv) cos 75°
Solution:
cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°
= \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

v) sin 105°
Solution:
sin 105° = sin (60° + 45°)
= sin 60° cos 45° + cos 60° sin 45°
= \(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

vi) tan 75°
Solution:
tan 75° = tan (45° + 30°)

vii) cot 75°
Solution:
cot 75° = \(\frac{1}{\tan 75^{\circ}}\)
= \(\frac{1}{2+\sqrt{3}}\)
= \(\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\)
= 2 + √3

viii) cos 105°
Solution:
cos 105° = cos (60° + 45°)
= cos 60° cos 45° – sin 60° sin 45°
= \(\frac{1}{2} \cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}\)
= \(\frac{1-\sqrt{3}}{2 \sqrt{2}}\)

ix) tan 105°
Solution:
tan 105° = tan (60° + 45°)

x) cot 105°
Solution:
cot 105° = \(\frac{1}{\tan 105^{\circ}}\)
= \(\frac{1}{-2-\sqrt{3}}\)
= \(-\frac{1}{2+\sqrt{3}}\)
= – (2 -√3)
= – 2 + √3

xi) cos 15°
Solution:
cos 15° = cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30°
= \(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

xii) sin 15°
Solution:
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
= \(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

Question 2.
Prove that sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x.
Solution:
= sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x
[∵ cos A cos B + sin A sin B = cos (A – B)]
= cos [(n + 2) x – (n + 1) x]
= cos [n x + 2 x – n x – x]
= cos x

II. Prove that

Question 1.
sin² \(\frac{\pi}{6}\) + cos² \(\frac{\pi}{3}\) – tan² \(\frac{\pi}{4}\) = – \(\frac{1}{2}\).
Solution:
L.H.S. = sin² \(\frac{\pi}{6}\) + cos² \(\frac{\pi}{3}\) – tan² \(\frac{\pi}{4}\)

Question 2.
2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\) = \(\frac{3}{2}\).
Solution:
L.H.S. = 2 sin² \(\frac{\pi}{6}\) + cosec² \(\frac{7 \pi}{6}\) cos² \(\frac{\pi}{3}\)

Question 3.
cot² \(\frac{\pi}{6}\) + cosec² \(\frac{5 \pi}{6}\) + 3 tan² \(\frac{\pi}{6}\) = 6.
Solution:
LHS = cot² \(\frac{\pi}{6}\) + cosec² \(\frac{5 \pi}{6}\) + 3 tan² \(\frac{\pi}{6}\)
= (√3)² + cosec (π – \(\frac{\pi}{6}\)) + 3 (\(\frac{1}{\sqrt{3}}\))²
= 3 + cosec \(\frac{\pi}{6}\) + 3 × \(\frac{1}{3}\)
= 3 + 2+ 1
= 6 = RHS

Question 4.
2 sin² \(\frac{3 \pi}{4}\) + 2 cos² \(\frac{\pi}{4}\) + 2 sec² \(\frac{\pi}{3}\) = 10.
Solution:
LHS = 2 sin² \(\frac{3 \pi}{4}\) + 2 cos² \(\frac{\pi}{4}\) + 2 sec² \(\frac{\pi}{3}\)
= 2 sin² (π – \(\frac{\pi}{4}\)) + 2 (\(\frac{1}{\sqrt{2}}\))² + 2 (2)²
= 2 × (\(\frac{1}{\sqrt{2}}\))² + 1 + 8
= 2 × ½ + 9
= 1 + 9
= 10 = RHS

Prove the following:

Question 5.
cos (\(\frac{\pi}{4}\) – x) cos (\(\frac{\pi}{4}\) – y) – sin (\(\frac{\pi}{4}\) – x) sin (\(\frac{\pi}{4}\)– y) = sin (x + y)
Solution:
LHS = cos (\(\frac{\pi}{4}\) – x) cos (\(\frac{\pi}{4}\) – y) – sin (\(\frac{\pi}{4}\) – x) sin (\(\frac{\pi}{4}\)– y)
= cos (\(\frac{\pi}{4}\) – x) + (latex]\frac{\pi}{4}[/latex] – y)]
[∵ cos A cos B – sin A sin B = cos (A + B)]
= cos [\(\frac{\pi}{2}\) – (x + y)]
= sin (x + y) = RHS

Question 6.
\(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^2\)
Solution:
\(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}\)

Question 7.
\(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\) = cot² x
Solution:
LHS = \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\)
= \(\frac{-\cos x \cos x}{\sin x(-\sin x)}\)
= \(\frac{-\cos ^2 x}{-\sin ^2 x}\)
= cot² x
= RHS

Question 8.
cos (\(\frac{3 \pi}{4}\) + x) – cos (\(\frac{3 \pi}{4}\) – x) = – √2 sin x.
Solution:
LHS = cos (\(\frac{3 \pi}{4}\) + x) – cos (\(\frac{3 \pi}{4}\) – x)

Question 9.
sin² 6x – sin² 4x = sin 2x sin 10x.
Solution:
LHS = sin² 6x – sin² 4x
= sin (6x + 4x) sin (6x – 4x)
[∵ sin² A – sin² B = sin (A + B) . sin (A – B)]
= sin 10x . sin 2x
= sin 2x sin 10x = RHS

Question 10.
cos² 2x – cos² 6x = sin 4x sin 8x
Solution:
LHS = cos² 2x – cos² 6x
= 1 – sin² 2x – (1 – sin² 6x)
= 1 – sin² 2x – 1 + sin² 6x
[∵ sin² A – sin² B = sin (A + B) . sin (A – B)]
= sin² 6x – sin² 2x
= sin (6x + 2x) sin (6x – 2x)
= sin 8x . sin 4x = RHS

Question 11.
\(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}\)
Solution:
LHS = \(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}\)

Question 12.
\(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\) = tan 4x
Solution:
LHS = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)

Question 13.
\(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}\)
Solution:
LHS = \(\frac{\sin x-\sin y}{\cos x+\cos y}\)

Question 14.
\(\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}\) = tan 2x
Solution:
LHS = \(\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}\)

Question 15.
\(\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}\) = 2 sin x
Solution:
LHS = \(\frac{\sin x-\sin 3 x}{\sin ^2 x-\cos ^2 x}\)

Question 16.
tan 4x = \(\frac{4 \tan x\left(1-\tan ^2 x\right)}{1-6 \tan ^2 x+\tan ^4 x}\).
Solution:
LHS = tan 4x
= tan 2 (2x)

Question 17.
cos 4x = 1 – 8 sin² x cos² x
Solution:
LHS = cos 4x
= cos 2 (2x)
(∵ cos 2x = 1 – 2 sin² x)
= 1 – 2 sin² x
= 1 – 2 (sin 2x)²
(∵ sin 2x = 2 sin x cos x)
= 1 – 2 ( 2 sin x cos x)²
= 1 – 2 (4 sin² x cos² x)
= 1 – 8 sin² x cos² x
= RHS

III. Prove the following:

Question 1.
cos (\(\frac{3 \pi}{2}\) + x) cos (2π + x) [cot (\(\frac{3 \pi}{2}\) – x) + cot (2π + x) = 1.
Solution:
LHS = cos (\(\frac{3 \pi}{2}\) + x) cos (2π + x) [cot (\(\frac{3 \pi}{2}\) – x) + cot (2π + x)
= sin x cos x [tan x + cot x]
[∵ cot (\(\frac{3 \pi}{2}\) – θ) = tan θ,
cos (\(\frac{3 \pi}{2}\) + θ) = sin θ]
= sin x cos x \(\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]\)
= sin x cos x \(\left[\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\right]\)
= sin² x + cos² x
= 1 = RHS

Question 2.
sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x.
Solution:
LHS = sin 2x + 2 sin 4x + sin 6x
= sin 6x + sin 2x + 2 sin 4x
= [2 sin \(\left(\frac{6 x+2 x}{2}\right)\) cos \(\left(\frac{6 x-2 x}{2}\right)\)] + 2 sin 4x
[∵ sin A + sin B = 2 sin \(\frac{A+B}{2}\) cos \(\frac{A-B}{2}\)]
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x . (cos 2x + 1)
= 2 sin 4x [(2 cos² x – 1]
= 4 cos² x sin 4x = RHS

Question 3.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – cos 3x)
Solution:
LHS = cot 4x (sin 5x + sin 3x)
= cot 4x [2 sin \(\left(\frac{5 x+3 x}{2}\right)\) cos \(\left(\frac{5 x-3 x}{2}\right)\)
[∵ sin A + sin B = 2 sin \(\frac{A+B}{2}\) cos \(\frac{A-B}{2}\)]
= \(\frac{\cos 4 x}{\sin 4 x}\) [2 sin 4x cos x]
= 2 cos 4x cos x ……………………..(1)
RHS = cot x (sin 5x – cos 3x)
= cot x [2 cos \(\left(\frac{5 x+3 x}{2}\right)\) sin \(\left(\frac{5 x-3 x}{2}\right)\)
[∵ sin A + sin B = 2 cos \(\frac{A+B}{2}\) sin \(\frac{A-B}{2}\)]
= \(\frac{\cos x}{\sin x}\) [2 cos 4x sin x]
= 2 cos 4x cos x …………………………..(2)
From equations (1) and (2),
LHS = RHS.

Question 4.
\(\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}\) = cot 3x
Solution:
LHS = \(\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}\)

Question 5.
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1.
Solution:
LHS = 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) (cot 2x + cot x).
= cot x cot 2x – \(\left(\frac{\cot 2x \cot x – 1}{\cot x + \cot 2x}\right)\) (cot 2x + cot x)
= cot x cot 2x – (cot 2x cot x – 1)
= cot x cot 2x – cot 2x cot x + 1
= 1 = R.H.S.

Question 6.
cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x
Solution:
cos 6x = cos 3(2x)
= 4 cos³ (2x) – 3 cos 2x
[∵ cos 3x = 4 cos³ x – 3 cos x]
= 4 (2 cos² x – 1)³ – 3(2 cos² x – 1)
[∵ cos 2x = 2 cos² x – 1]
= 4 [(2 cos² x)³ (- 1)³ – 3 (2 cos² x)² + 3 (2 cos² x)] – 6 cos² x + 3
= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1
Hence proved.