Question

Find the value of
${sin}^{ - 1} (sin( - \frac{\pi}{6} ))$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given inverse Trigonometric function is

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg( - \dfrac{\pi}{6} \bigg) \bigg)$

We know,

$\boxed{ \bf{sin( - x) \: = \: - \: sinx }}$

So, using this,

$\rm \: \: = \:{sin}^{ - 1} \bigg( - sin \bigg(\dfrac{\pi}{6} \bigg) \bigg)$

We know,

$\boxed{ \bf{sin^{ - 1} ( - x) \: = \: - \: sin ^{ - 1} x }}$

So, using this

$\rm \: \: = - \:{sin}^{ - 1} \bigg(sin \bigg(\dfrac{\pi}{6} \bigg) \bigg)$

Now, we further know that

$\boxed{ \bf{ {sin}^{ - 1} (sinx) = x \: \: if \: x \in \: \bigg[ - \dfrac{\pi}{2} , \: \dfrac{\pi}{2} \bigg] \: }}$

So, using this,

$\rm \: \: = \: - \: \dfrac{\pi}{6}$

Hence,

$\bf\implies \:\:{sin}^{ - 1} \bigg( sin \bigg( - \dfrac{\pi}{6} \bigg) \bigg) = - \: \dfrac{\pi}{6}$

Additional Information :-

Let us assume one more example.

Evaluate the following :-

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg(\dfrac{7\pi}{6} \bigg) \bigg)$

We know,

$\boxed{ \bf{ {sin}^{ - 1} (sinx) = x \: \: if \: x \in \: \bigg[ - \dfrac{\pi}{2} , \: \dfrac{\pi}{2} \bigg] \: }}$

But

$\boxed{ \bf{ \dfrac{7\pi}{6} \: \: \cancel \in \: \bigg[ - \dfrac{\pi}{2} , \: \dfrac{\pi}{2} \bigg] \: }}$

So,

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg(\dfrac{7\pi}{6} \bigg) \bigg)$

can be rewritten as

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg(\pi + \dfrac{\pi}{6} \bigg) \bigg)$

$\rm \: = \:\:{sin}^{ - 1} \bigg( - sin \bigg(\dfrac{\pi}{6} \bigg) \bigg)$

$\rm \: = \:\: - \: {sin}^{ - 1} \bigg(sin \bigg(\dfrac{\pi}{6} \bigg) \bigg)$

$\rm \: \: = \: - \: \dfrac{\pi}{6}$

Hence,

$\bf\implies \:\:{sin}^{ - 1} \bigg( sin \bigg( \dfrac{7\pi}{6} \bigg) \bigg) = - \: \dfrac{\pi}{6}$

Let take one more example

Evaluate the following :-

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg(\dfrac{5\pi}{6} \bigg) \bigg)$

We know,

$\boxed{ \bf{ {sin}^{ - 1} (sinx) = x \: \: if \: x \in \: \bigg[ - \dfrac{\pi}{2} , \: \dfrac{\pi}{2} \bigg] \: }}$

But

$\boxed{ \bf{ \dfrac{5\pi}{6} \: \: \cancel \in \: \bigg[ - \dfrac{\pi}{2} , \: \dfrac{\pi}{2} \bigg] \: }}$

So,

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg(\dfrac{5\pi}{6} \bigg) \bigg)$

can be rewritten as

$\rm :\longmapsto\:\:{sin}^{ - 1} \bigg(sin \bigg(\pi - \dfrac{\pi}{6} \bigg) \bigg)$

$\rm \: = \:\:{sin}^{ - 1} \bigg(sin \bigg(\dfrac{\pi}{6} \bigg) \bigg)$

$\rm \: \: = \: \: \dfrac{\pi}{6}$

Hence,

$\bf\implies \:\:{sin}^{ - 1} \bigg( sin \bigg( \dfrac{5\pi}{6} \bigg) \bigg) = \: \dfrac{\pi}{6}$