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

Free Inverse Functions Calculator – find functions inverse step-by-step.

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) -10π ≤ x ≤ 10π and sin x = 0

sin x = 0, sin θ = sin α

sin x = sin 0, θ = nπ + (-1)^{n} a, n ∈ R

x = nπ + (-1)^{n} (0)

x = nπ, n = 0, ± 1, ±2, …….. ± 10

n = ± 1, ±2, … ± 10

(ii) -3π ≤ x ≤ 3π and sin x = -1.

sin x = -1

sin x = -sin \(\frac{π}{2}\) = sin(-\(\frac{π}{2}\))

x = (4n – 1)\(\frac{π}{2}\), n = 0, ± 1

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.

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.

The graph is

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:

Question 5.

For that value of x does sin x = sin^{-1} x?

Solution:

Let y = sin^{-1} x

When y = 0 ⇒ 0 = sin^{-1} x

sin 0 = sin [sin^{-1}(x)]

sin 0 = x

∴ x = 0

Only when x = 0, then sin x = sin^{-1}(x)

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 \(\frac{x^{2}+1}{2 x} \leq 1\)

⇒ x^{2} + 1 ≥ -2x or x^{2} + 1 ≤ 2x

⇒ x^{2} + 1 + 2x ≥ 0 or x^{2} + 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]

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 Additional Problems

Question 1.

Solution:

Question 2.

Solution: