Inverse Trignometric Function Examples

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

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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.

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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)

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

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

= $\pi$

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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.