AP Inter 1st Year Maths Exercise 3d Solutions

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

AP Inter 1st Year Maths Trigonometric Functions Exercise 3d

I. Find sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) in each of the following.

Question 1.
tan = – \(\frac{4}{3}\), x in quadrant II.
Solution:
Given, tan x = – \(\frac{4}{3}\) (x is in second quadrant)
i.e. \(\frac{\pi}{2}\) < x < π
⇒ \(\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}\)
i.e. \(\frac{x}{2}\) lies in the I quadrant,
so that all trigonometric ratios of \(\frac{x}{2}\) are +ve.
Now, sec² x = 1 + tan² x
= 1 + \(\frac{16}{9}\)
= \(\frac{9+16}{9}\)
= \(\frac{25}{9}\)
∴ sec x = ± \(\frac{5}{3}\)
But x is in II quadrant
∵ sec x is -ve. i.e., sec x = – \(\frac{5}{3}\)
⇒ cos x = – \(\frac{3}{5}\)

Question 2.
cos x = – \(\frac{1}{3}\), x in quadrant II.
Solution:
Given, cos x = – \(\frac{1}{3}\) [x is in quadrant II]
i.e., π < x < \(\frac{3 \pi}{2}\)
⇒ \(\frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}\)
⇒ 90° < \(\frac{x}{2}\) < 135°
i.e., \(\frac{x}{2}\) lies in II quadrant, so that sin \(\frac{x}{2}\) > 0, cos \(\frac{x}{2}\) < 0 and tan \(\frac{x}{2}\) < 0

Question 3.
sin x = \(\frac{1}{4}\), x in quadrant II.
Solution:
Given, sin x = \(\frac{1}{4}\), [x is in quadrant II]
i.e. \(\frac{\pi}{2}\) < x < π
⇒ \(\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}\)
i.e., \(\frac{x}{2}\) lies in I quadrant,
so that all trigonometric ratios of \(\frac{x}{2}\) are +ve.
Also, cos² x = 1 – sin² x
= 1 – \(\frac{1}{16}\) = \(\frac{15}{16}\)
cos x = ± \(\frac{\sqrt{15}}{4}\)
But x is in II quadrant and cos x < 0.
∴ cos x = \(\frac{-\sqrt{15}}{4}\)

II. Prove the following.

Question 1.
2 cos \(\frac{\pi}{13}\) cos \(\frac{9 \pi}{13}\) + cos \(\frac{3 \pi}{13}\) + cos \(\frac{5 \pi}{13}\) = 0
Solution:
LHS = 2 cos \(\frac{\pi}{13}\) cos \(\frac{9 \pi}{13}\) + cos \(\frac{3 \pi}{13}\) + cos \(\frac{5 \pi}{13}\)

Question 2.
(sin 3x + sin x) sinx + (cos 3x – cos x) cos x = 0.
Solution:
LHS = (sin 3x + sin x) sinx + (cos 3x – cos x) cos x
= sin 3x sin x + sin² x + cos 3x cos x – cos² x
= cos 3x cos x + sin 3x sin x – (cos² x – sin² x)
= cos (3x – x) – cos 2x
[∵ cos (A – B) = cos A cos B + sin A sin B]
= cos 2x – cos 2x = 0 = RHS

Question 3.
(cos x + cos y)² + (sin x – sin y)² = 4 cos² \(\left(\frac{x+y}{2}\right)\)
Solution:
LHS = (cos x + cosy)² + (sin x – sin y)²
= cos² x + cos² y + 2 cos x . cos y + sin² x + sin² y – 2 sin x . sin y
= cos² x + sin² x + cos² y + sin² y + 2(cos x . cos y – sin x . sin y)
= 1 + 1 + 2 cos (x + y)
= 2 + 2 cos (x + y)
= 2 (1 + cos(x + y))
= 2 . 2 cos² \(\left[\frac{\mathrm{x}+\mathrm{y}}{2}\right]\)
= 4 cos² \(\left[\frac{\mathrm{x}+\mathrm{y}}{2}\right]\) = RHS

Question 4.
(cos x – cos y)² + (sin x – sin y)² = 4 sin²
Solution:
LHS = (cos x – cos y)² + (sin x – sin y)²
= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y – 2sin x sin y
= (cos² x + sin² x) + (cos² y + sin² y) – 2 [cos x cos y + sin x sin y]
= 1 + 1 – 2 [cos (x – y)]
[∵ cos (A – B) = cos A cos B + sin A sin B]
= 2 [1 – cos (x – y)]
= 2 [2 sin² \(\left[\frac{\mathrm{x}-\mathrm{y}}{2}\right]\)]
= 4 sin² \(\left[\frac{\mathrm{x}-\mathrm{y}}{2}\right]\) = RHS

Question 5.
sin 3x + sin 2x – sin x = 4 sin x cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\)
Solution:
L.H.S. = sin 3x + sin 2x – sin x
= sin 3x – sin x + sin 2x
= 2 cos \(\left(\frac{3 x+x}{2}\right)\) sin \(\left(\frac{3 x-x}{2}\right)\) + 2 sin x cos x
= 2 cos 2x sin x + 2 sin x cos x
= 2 sin x (cos 2x + cos x)
= 4 sinx cos \(\frac{3 x}{2}\) cos \(\frac{x}{2}\)
= 4 sin x cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\) = RHS

III. Prove the following.

Question 1.
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x.
Solution:
LHS = sin x + sin 3x + sin 5x + sin 7x
= (sin x + sin 5x) + (sin 3x + sin 7x)
= 2 sin \(\left[\frac{x+5 x}{2}\right]\) cos \(\left[\frac{x-5 x}{2}\right]\) + 2 sin \(\left[\frac{3 x+7 x}{2}\right]\) cos \(\left[\frac{3 x-7 x}{2}\right]\)
= 2 sin 3x cos (- 2x) + 2 sin 5x cos (- 2x)
= 2 sin 3x cos 2x + 2 sin 5x cos 2x
= 2 cos 2x [sin 3x + sin 5x]
= 2 cos 2x [2 sin \(\left[\frac{3 x+5 x}{2}\right]\) . cos \(\left[\frac{3 x-5 x}{2}\right]\)]
= 2 cos 2x [2 sin 4x . cos (- x)]
= 4 cos 2x sin 4x cos x = RHS

Question 2.
\(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x
Solution:
\(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\)