Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.1 Textbook Questions and Answers, Notes.

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

Question 1.

Find all the values of x such that

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

Solution:

(i) y = sin 7x

For Amplitude use the form

y = a sin(bx – c) + d

Amplitude = |a|

a = 1, ∴ |a| = 1

Period using the formula = \(\frac{2π}{|b|}\) = \(\frac{2π}{|7|}\) = \(\frac{2π}{7}\)

∴ Amplitude = 1

Period = \(\frac{2π}{7}\)

(ii) y = -sin (\(\frac{1}{3}\)x)

a = -1, b = \(\frac{1}{3}\)

∴ Amplitude = |a| = |-1| = 1

Period = \(\frac{2π}{|b|}\) = \(\frac{2π}{| \frac{1}{3}|}\) = 3(2π) = 6π

(iii) y =4 sin (-2x)

a = 4, b = -2

Amplitude = |4| = 4

Period = \(\frac{2π}{|b|}\) = \(\frac{2π}{|-2|}\) = \(\frac{2π}{2}\) = π

Question 3.

Sketch the graph of y = sin (\(\frac{1}{3}\)x) for 0 ≤ x ≤ 6π

Solution:

Question 4.

Find the value of

(i) sin^{-1} (sin(\(\frac{2π}{3}\)))

(ii) sin^{-1} (sin(\(\frac{5π}{4}\)))

Solution:

Question 5.

For what 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} (\(\frac{x²+1}{2x}\))

Solution:

f(x) = sin^{-1} (\(\frac{x²+1}{2x}\))

Range of sin^{-1}x is [-1, 1]

-1 ≤ \(\frac{x²+1}{2x}\) ≤ 1

Multiply by 2x² ≥ 0

∴ solution is [-1, 1]

(ii) g (x) = 2 sin^{-1} (2x – 1) – \(\frac{π}{4}\)

Range of 2 sin^{-1} x is [-1, 1]

-1 ≤ 2x – 1 ≤ 1

2x – 1 ≤ 1 (or) 2x – 1 ≥ -1

2x ≤ 2 (or) 2x ≥ -1 + 1

x ≤ 1 (or) 2x ≥ 0 ⇒ x ≥ 0

x ≥ 0 and x ≤ 1

∴ solution is [0, 1]

Question 7.

Find the value of

(sin\(\frac{5π}{9}\) cos\(\frac{π}{9}\) + cos\(\frac{5π}{9}\) sin\(\frac{π}{9}\))

Solution:

sin (A + B) = sin A cos B + cos A sin B