Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1

Question 1.
Find all the values of x such that
(i) -10π ≤ x ≤ 10π and sin x = 0
(ii) -8π ≤ x ≤ 8π and sin x = -1
Solution:
(i) sin x = 0
⇒ x = nπ
where n = 0, ±1, ±2, ±3, ……., ±10
(ii) sin x = -1
⇒ x = (4n – 1) \(\frac{\pi}{2}\), n = 0, ±1, ±2, ±3, 4

Question 2.
Find the period and amplitude of
(i) y = sin 7x
(ii) y = -sin(\(\frac{1}{3}\)x)
(iii) y = 4 sin(-2x)
Solution:
(i) y = sin 7x
Period of the function sin x is 2π
Period of the function sin 7x is \(\frac{2 \pi}{7}\)
The amplitude of sin 7x is 1.

(ii) y = -sin\(\frac{1}{3}\)x
Period of sin x is 2π
So, period of sin\(\frac{1}{3}\)x is 6π and the amplitude is 1.

(iii) y = 4 sin(-2x) = -4 sin 2x
Period of sin x is 2π
π Period of sin 2x is π and the amplitude is 4.

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1

Question 3.
Sketch the graph of y = sin(\(\frac{1}{3}\)x) for 0 ≤ x < 6π.
Solution:
The period of sin(\(\frac{1}{3}\)x) is 6π and the amplitude is 1.
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 Q3
The graph is
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 Q3.1

Question 4.
Find the value of
(i) \(\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)\)
(ii) \(\sin ^{-1}\left(\sin \left(\frac{5 \pi}{4}\right)\right)\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 Q4

Question 5.
For that value of x does sin x = sin-1 x?
Solution:
sin x = sin-1 x is possible only when x = 0 (∵ x ∈ R)

Question 6.
Find the domain of the following
(i) \(f(x)=\sin ^{-1}\left(\frac{x^{2}+1}{2 x}\right)\)
(ii) \(g(x)=2 \sin ^{-1}(2 x-1)-\frac{\pi}{4}\)
Solution:
(i) \(f(x)=\sin ^{-1}\left(\frac{x^{2}+1}{2 x}\right)\)
The range of sin-1 x is -1 to 1
\(-1 \leq \frac{x^{2}+1}{2 x} \leq 1\)
⇒ \(\frac{x^{2}+1}{2 x} \geq-1\) or \(\frac{x^{2}+1}{2 x} \leq 1\)
⇒ x2 + 1 ≥ -2x or x2 + 1 ≤ 2x
⇒ x2 + 1 + 2x ≥ 0 or x2 + 1 – 2x ≤ 0
⇒ (x+ 1)2 ≥ 0 or (x – 1)2 ≤ 0 which is not possible
⇒ -1 ≤ x ≤ 1 or
(ii) \(g(x)=2 \sin ^{-1}(2 x-1)-\frac{\pi}{4}\)
-1 ≤ (2x – 1) ≤ 1
0 ≤ 2x ≤ 2
0 ≤ x ≤ 1
x ∈ [0, 1]

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1

Question 7.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 Q7

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 Additional Problems

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 1
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 2

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 3
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 4

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