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

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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: