Question

# Find the value of ${\sin ^{ - 1}}\{ \sin ( - {600^ \circ })\}$

Hint: To solve this problem we need to have basic knowledge about the trigonometric values, trigonometric identities and inverse trigonometric identities because the question here belongs to the inverse trigonometry concept.

Before solving this problem let us consider the given term as P.
Then $P = {\sin ^{ - 1}}\{ \sin ( - {600^ \circ })\}$
On using the trigonometric identity $\sin ( - \theta ) = - \sin \theta$ we can rewrite P as
$\Rightarrow {\sin ^{ - 1}}\{ - \sin ({600^ \circ })\}$
$\Rightarrow {\sin ^{ - 1}}\{ - \sin ({360^ \circ } \times 2 - {120^ \circ })\}$ $[\because {360^ \circ } = 2\pi ]$
$\Rightarrow {\sin ^{ - 1}}\{ - \sin (2\pi - {120^ \circ })\}$
Now by using the trigonometric identity $\sin (2\pi - A) = \sin ( - A)$ we can rewrite the above term as
$\Rightarrow {\sin ^{ - 1}}\{ - \sin ( - {120^ \circ })\}$
Again by using the trigonometric identity $\sin ( - \theta ) = - \sin \theta$ we can rewrite the term as $\Rightarrow {\sin ^{ - 1}}\{ - ( - \sin {120^ \circ })\} \\ \Rightarrow {\sin ^{ - 1}}\{ \sin ({120^ \circ })\} \\$
Now this can also be written as
$\Rightarrow {\sin ^{ - 1}}\{ \sin ({180^ \circ } - {60^ \circ })\}$
$\Rightarrow {\sin ^{ - 1}}\{ \sin ({60^ \circ })\}$
On using the inverse trigonometric identity ${\sin ^{ - 1}}\{ \sin x\} = x$ we can rewrite the term as
$\Rightarrow {60^ \circ } \\ \therefore P = {60^ \circ } \\$
Hence the value of ${\sin ^{ - 1}}\{ \sin ( - {600^ \circ })\} = {60^ \circ }$

Note: The above solution is a step-by-step process of finding the value of a given term where we have included the trigonometric identities and inverse trigonometric identities to solve the question. This can also be done in a simple way, as $\sin ( - {600^ \circ }) = \sin ( - {600^ \circ } + {720^ \circ }) = \sin ({120^ \circ }) = \sin ({60^ \circ })$
Here finally we have found the theta value so the answer is $\theta = {60^ \circ }$.