Inter 1st Year Maths Trigonometric Functions Solutions Exercise 3d

Practicing the AP Board Solutions Class 11 Maths and Chapter 3 Inter 1st Year Maths Trigonometric Functions Solutions Exercise 3d Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Trigonometric Functions Solutions Exercise 3d

Trigonometric Functions Exercise 3d Solutions

Trigonometric Functions Class 11 Exercise 3d Solutions – Trigonometric Functions 3d Exercise Solutions

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

Question 1.
tan x = \(-\frac {4}{3}\), x in quadrant II
Solution:

Question 2.
cos x = \(-\frac {1}{3}\), x in quadrant III
Solution:

Question 3.
sin x = \(\frac {1}{4}\), x in quadrant II
Solution:

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:

Question 2.
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Solution:
Consider
L.H.S = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
By further calculation
sin 3x sin x + sin²x + cos 3x cos x – cos²x
Taking out the common terms
cos 3x cos x + sin 3x sin x – (cos²x – sin²x)
Using the formula cos (A – B) = cos A cos B + sin A sin B
cos (3x – x) – cos 2x
So, we get
cos 2x – cos 2x = 0 = R.H.S

Question 3.
(cos x + cos y)² + (sin x – sin y)² = \(4 \cos ^2 \frac{x+y}{2}\)
Solution:
Consider
L.H.S = (cos x + cos y)² + (sin x – sin y)²
By expanding using the formula, we get
cos²x + cos²y + 2 cos x cos y + sin²x + sin²y – 2 sin x sin y
Grouping the terms
(cos²x + sin²x) + (cos²y + sin²y) + 2(cos x cos y – sin x sin y)
Using the formula
cos(A + B) – (cos A cos B – sin A sin B)
1 + 1 + 2 cos(x + y)
By further calculation
2 + 2 cos(x + y)
Taking 2 as common
2[1 + cos(x + y)]
from the formula cos 2A = 2 cos²A – 1
\(2\left[1+2 \cos ^2\left(\frac{x+y}{2}\right)-1\right]\)
We get \(4 \cos ^2\left(\frac{x+y}{2}\right)\) = R.H.S

Question 4.
(cos x – cos y)² + (sin x – sin y)² = \(4 \sin ^2 \frac{x-y}{2}\)
Solution:
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
= (cos²x + sin²x) + (cos²y + sin²y) – 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)]
= \(2\left[2 \sin ^2\left[\frac{x-y}{2}\right]\right]\)
= \(4 \sin ^2\left[\frac{x-y}{2}\right]\)
= R.H.S

Question 5.
sin 3x + sin 2x – sin x = 4 sin x cos \(\frac {x}{2}\) cos \(\frac {3x}{2}\)
Solution:
L.H.S = sin 3x + sin 2x – sin x
It can be written as
sin 3x + (sin 2x – sin x)
Using the formula

III. Prove the following:

Question 1.
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
Solution:
Consider
L.H.S = sin x + sin 3x + sin 5x + sin 7x
Grouping the terms
(sin x + sin 5x) + (sin 3x + sin 7x)
Using the formula

By further calculation
2 sin 3x cos (-2x) + 2 sin 5x cos (-2x)
we get 2 sin 3x cos 2x + 2 sin 5x cos 2x
Taking out the common terms
2 cos 2x [sin 3x + sin 5x]
Using the formula, we can write it as
\(2 \cos 2 x \cdot\left[2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)\right]\)
we get 2 cos 2x [2 sin (4x . cos (-x)]
= 4 cos x cos 2x sin 4x
= R.H.S.

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: