Question

How do you solve $\sin x = 0.25$ ?

Answer 1

Hint : In order to find the solution of a trigonometric equation, we start by taking the inverse trigonometric function like inverse sin, inverse cosine, inverse tangent on both sides of the equation and then set up reference angles to find the rest of the answers.
For ${\sin ^{ - 1}}$ function, the principal value branch is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
For ${\cos ^{ - 1}}$ function, the principal value branch is $\left[ {0,\pi } \right]$ .
For ${\tan ^{ - 1}}$ function, the principal value branch is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ .

Complete step by step solution:
According to definition of inverse ratio,
If $\sin x = 0.25$ ,
Then, $x = {\sin ^{ - 1}}\left( {0.25} \right)$ where the value of x lies in the range $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
Now, we know that the sine function is positive in the first and second quadrants and negative in the fourth and third quadrant.
So, the angle x must originally lie either in the first quadrant or second quadrant. But the value of $x = {\sin ^{ - 1}}\left( {0.25} \right)$ will lie in the range $\left[ {0,\dfrac{\pi }{2}} \right]$ .
Simplifying the expression further, we get,
$\Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{1}{4}} \right)$
But there is no standard angle for which $\sin x = 0.25$ . But, we know that the value of $\sin \left( {{{15}^ \circ }} \right)$ is slightly greater than $0.25$ . So, the value of angle x for which $\sin x = 0.25$ is approximately ${14.5^ \circ }$ .
Hence, the solution of $\sin x = 0.25$ is ${14.5^ \circ }$ approximately.
So, the correct answer is “ $\sin x = 0.25$ is ${14.5^ \circ }$ ”.

Note : The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of the right-angled triangles. The value of $\sin \left( {{{15}^ \circ }} \right)$ can be calculated using half angle formula of cosine $\cos \left( {2\theta } \right) = 1 - 2{\sin ^2}\theta$ . So, value of $\sin \left( {{{15}^ \circ }} \right)$ can be calculated as $\cos \left( {{{30}^ \circ }} \right) = 1 - 2{\sin ^2}\left( {{{15}^ \circ }} \right)$ .