Question

If $\sin \left( {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x \right)=1$ , then find the value of x.

Answer 1

Hint: At first write the value of 1 as $\sin \dfrac{\pi }{2}$ . Then compare the expression to write expression of ${{\sin }^{-1}}\left( \dfrac{1}{5} \right)+{{\cos }^{-1}}x$ as \[\dfrac{\pi }{2}\] . Then rewrite the given expression as ${{\sin }^{-1}}\left( \dfrac{1}{5} \right)={{\sin }^{-1}}x$ . Hence find x.

Complete step-by-step answer:
In the question we are given an equation $\sin \left( {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x \right)=1$ and we have to solve and find the value of x.

Before proceeding let us know what are inverse trigonometric functions. Inverse trigonometric functions are functions that are inverse functions of trigonometric functions. Specifically they are inverses of sine, cosine, tangent, cotangent, secant, cosecant functions, and are used to obtain an angle from any of the angles' trigonometric ratios.

These are certain notations which are used. Some of the most common notation is using arc $\sin \left( x \right)$ , arc $\cos \left( x \right)$ , arc $\tan \left( x \right)$ instead of ${{\sin }^{-1}}\left( x \right)$ , ${{\cos }^{-1}}\left( x \right)$ and ${{\tan }^{-1}}\left( x \right)$ . These arise from geometric relationships. When measuring in radius, an angle $\theta $ radius will correspond to an arc whose length is r $\theta $ , where r is radius of circle. Thus in the unit circle, “the arc whose cosine is x” is the same as “the angle whose cosine is x”, and the length of the arc of the circle in radii is the same as the measurement of angle in radius.

So we are given equation as –

 $\sin \left( {{\sin }^{-1}}\dfrac{1}{5}+{{\cos }^{-1}}x \right)=1$ .

We know that the value of $\sin \dfrac{\pi }{2}=1$ .So we can rewrite the equation as –

$\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{5} \right)+{{\cos }^{-1}}x \right)=\sin \dfrac{\pi }{2}$ .

So on comparing we can say that the value of expression ${{\sin }^{-1}}\left( \dfrac{1}{5}

\right)+{{\cos }^{-1}}x$ is $\dfrac{\pi }{2}$ .

So, we can write it as,

${{\sin }^{-1}}\left( \dfrac{1}{5} \right)+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ .

We know a identity that ${{\sin }^{-1}}\left( x \right)+{{\cos }^{-1}}\left( x \right)$ is equal to $\dfrac{\pi }{2}$. So, we can write ${{\cos }^{-1}}x$ as $\left( \dfrac{\pi }{2}-{{\sin }^{-1}}x \right)$. Hence we can write equation as –

${{\sin }^{-1}}\left( \dfrac{1}{5} \right)+\dfrac{\pi }{2}-{{\sin }^{-1}}x=\dfrac{\pi }{2}$ .

Hence, on simplification we can write the equation as –

${{\sin }^{-1}}\left( \dfrac{1}{5} \right)={{\sin }^{-1}}\left( x \right)$ .

Now on comparing we can say that the value of x is $\dfrac{1}{5}$ .

So the answer is $\dfrac{1}{5}$.



Note: We can do this problem just by visualizing. We know that the value of $\sin \dfrac{\pi }{2}$ is 1, so, we can say that the value of ${{\sin }^{-1}}\left( \dfrac{1}{5} \right)+{{\cos }^{-1}}x$ is $\dfrac{\pi }{2}$ . We know the identity ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ . So, on comparing we can say that x is also $\dfrac{1}{5}$.