# Inverse Trignometric Function Examples

Here you will learn some inverse trignometric function examples for better understanding of inverse trigonometric function concepts.

Example 1 : Find the value of $$sin^{-1}({-\sqrt{3}\over 2})$$ + $$cos^{-1}(cos({7\pi\over 6}))$$.

Solution : $$sin^{-1}({-\sqrt{3}\over 2})$$ = – $$sin^{-1}({\sqrt{3}\over 2})$$ = $$-\pi\over 3$$

$$cos^{-1}(cos({7\pi\over 6}))$$ = $$cos^{-1}(cos({2\pi – {5\pi\over 6}}))$$ = $$cos^{-1}(cos({5\pi\over 6}))$$ = $$5\pi\over 6$$

Hence $$sin^{-1}({-\sqrt{3}\over 2})$$ + $$cos^{-1}(cos({7\pi\over 6}))$$ = $$-\pi\over 3$$ + $$5\pi\over 6$$ = $$\pi\over 2$$

Example 2 : Prove that : $$cos^{-1}{12\over 13}$$ + $$sin^{-1}{3\over 5}$$ = $$sin^{-1}{56\over 65}$$

Solution : We have, L.H.S. = $$cos^{-1}{12\over 13}$$ + $$sin^{-1}{3\over 5}$$ = $$tan^{-1}{5\over 12}$$ + $$tan^{-1}{3\over 4}$$

$$\because$$   [ $$cos^{-1}{12\over 13}$$ = $$tan^{-1}{5\over 12}$$ & $$sin^{-1}{3\over 5}$$ = $$tan^{-1}{3\over 4}$$ ]

L.H.S. = $$tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})$$ = $$tan^{-1}{56\over 33}$$

R.H.S. = $$sin^{-1}{56\over 65}$$ = $$tan^{-1}{56\over 33}$$

L.H.S = R.H.S.   Hence Proved.

Example 3 : Evaluate $$sin^{-1}(sin10)$$

Solution : We know that $$sin^{-1}(sinx)$$ = x, if $$-\pi\over 2$$ $$\le$$ x $$\le$$ $$\pi\over 2$$

Here, x = 10 radians which does not lie between -$$\pi\over 2$$ and $$\pi\over 2$$

But, $$3\pi$$ – x i.e. $$3\pi$$ – 10 lie between -$$\pi\over 2$$ and $$\pi\over 2$$

Also, sin($$3\pi$$ – 10) = sin 10

$$\therefore$$   $$sin^{-1}(sin10)$$ = $$sin^{-1}(sin(3\pi – 10)$$ = ($$3\pi$$ – 10)

Example 4 : Prove that : $$sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$

Solution : We have, A = $$sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$

A = $$tan^{-1}{12\over 5}$$ + $$tan^{-1}{3\over 4}$$ + $$tan^{-1}{63\over 16}$$

$$\implies$$ A = $$\pi$$ + $$tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})$$ + $$tan^{-1}{63\over 16}$$

$$\implies$$ A = $$\pi$$ + $$tan^{-1}{63\over (-16)}$$ + $$tan^{-1}{63\over 16}$$

= $$\pi$$ – $$tan^{-1}{63\over 16}$$ + $$tan^{-1}{63\over 16}$$

= $$\pi$$

Example 5 : Solve the equation : 2$$tan^{-1}({2x+1})$$ = $$cos^{-1}x$$

Solution : Here, 2$$tan^{-1}({2x+1})$$ = $$cos^{-1}x$$

cos(2$$tan^{-1}({2x+1})$$) = x       { We Know cos2x = $${1-tan^2x\over {1+tan^2x}}$$}

$$\therefore$$     $${{1-{(2x+1)}^2}\over {1-{(2x+1)}^2}}$$ = x   $$\implies$$   (1 – 2x – 1)(1 + 2x + 1) = x($$4x^2 + 4x + 2$$)

$$\implies$$   -2x.2(x + 1) = 2x($$2x^2 + 2x + 1$$)   $$\implies$$   2x($$2x^2 + 2x + 1 + 2x + 2$$) = 0

$$\implies$$   x = 0   or   $$2x^2 + 4x + 3$$ = 0   { No Solution }

Verify       x = 0

$$2tan^{-1}(1)$$ = $$cos^{-1}(1)$$   $$\implies$$   $$\pi\over 2$$ = $$\pi\over 2$$

$$\therefore$$     x = 0 is only the solution.

Practice these given inverse trignometric function examples to test your knowledge on concepts of inverse trigonometric function.