# Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.2

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.2

Question 1.
Find all the values of x such that
(i) -6π ≤ x ≤ 6π and cos x = 0
(ii) -5π ≤ x ≤ 5π and cos x = 1
Solution:
(i) cos x = 0
⇒ x = (2n + 1) ± $$\frac{\pi}{2}$$
n = 0, ±1, ±2, ±3, ±4, ±5
(ii) cos x = 1 = cos 0
⇒ x = 2nπ ± 0
n = 0, ±1, ±2

Question 2.
State the reason for $$\cos ^{-1}\left[\cos \left(-\frac{\pi}{6}\right)\right] \neq-\frac{\pi}{6}$$
Solution:
We know cos(-π) = cos π
so

Question 3.
Is cos-1 (-x) = π – cos-1 x true? Justify your answer.
Solution:
cos-1 (-x) = π – cos-1 x
Take x = cos θ ⇒ θ = cos-1 x
cos-1 (-x) = π – θ
cos(π – θ) = -cosθ = -x ∈ [-1, 1]
π – cos-1x = π – θ
π – cos-1 x = cos-1 (-x) is true

Question 4.
Find the principal value of $$\cos ^{-1}\left(\frac{1}{2}\right)$$
Solution:
Let $$\cos ^{-1}\left(\frac{1}{2}\right)$$ = π
⇒ cos π = $$\frac { 1 }{ 2 }$$ = cos $$\frac{\pi}{3}$$
⇒ π = $$\frac{\pi}{3}$$
⇒ $$\cos ^{-1}\left(\frac{1}{2}\right)$$ = $$\frac{\pi}{3}$$

Question 5.
Find the value of
(i) $$2 \cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)$$
(ii) $$\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}(-1)$$
(iii) $$\cos ^{-1}\left(\cos \frac{\pi}{7} \cos \frac{\pi}{17}-\sin \frac{\pi}{7} \sin \frac{\pi}{17}\right)$$
Solution:

Question 6.
Find the domain of
(i) $$f(x)=\sin ^{-1}\left(\frac{|x|-2}{3}\right)+\cos ^{-1}\left(\frac{1-|x|}{4}\right)$$
(ii) g(x) = sin-1 x + cos-1 x
Solution:
(i) $$f(x)=\sin ^{-1}\left(\frac{|x|-2}{3}\right)+\cos ^{-1}\left(\frac{1-|x|}{4}\right)$$ = U(x) + V(x) say
U(x):
$$-1<\frac{|x|-2}{3}<1$$
-3 < |x| – 2 < 3
-1 < |x| ≤ 5
V(x):
$$-1 \leq \frac{1-|x|}{4} \leq 1$$
-4 ≤ 1 – |x| ≤ 4
-5 ≤ -|x| ≤ 3
-3 ≤ |x| ≤ 5
from U(x) and V(x)
⇒ |x| ≤ 5
⇒ -5 ≤ |x| ≤ 5
(ii) g(x) = sin-1 x + cos-1 x
-1 ≤ x ≤ 1

Question 7.
For what value of x, the inequality $$\frac{\pi}{2}$$ < cos-1 (3x -1) < π holds?
Solution:
$$\frac{\pi}{2}$$ < cos-1 (3x -1) < π
⇒ cos$$\frac{\pi}{2}$$ < 3x – 1 < cos π
⇒ 0 < 3x – 1 < -1
⇒ 1 < 3x < 0
⇒ $$\frac{1}{3}$$ < x < 0
This inequality holds only if x < 0 or x > $$\frac{1}{3}$$

Question 8.
Find the value of
(i) $$\cos \left(\cos ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)\right)$$
(ii) $$\cos ^{-1}\left(\cos \left(\frac{4 \pi}{3}\right)\right)+\cos ^{-1}\left(\cos \left(\frac{5 \pi}{4}\right)\right)$$
Solution:

### Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.2 Additional Problems

Question 1.
Find the principal value of:
Solution:

Question 2.

Solution:

Question 3.

Solution: