Question

If $4{{\sin }^{-1}}x+{{\cos }^{-1}}x=\pi $, then find the value of x?
(a) 2
(b) 1
(c) $\dfrac{1}{3}$
(d) $\dfrac{1}{2}$

Answer 1

Hint:First, before proceeding for this, we must know the basic formula of the trigonometric relationship between the sine and cosine function as ${{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x$. Then, the value of ${{\cos }^{-1}}x$ found above in the given expression, we get the required form. Then, by taking the sine function on both sides to get the function only in x terms, we get the desired value of x as required in the question which a value of the sine function.

Complete step by step answer:
In this question, we are supposed to find the value of x when the expression is given as $4{{\sin }^{-1}}x+{{\cos }^{-1}}x=\pi $.
So, before proceeding for this, we must know the basic formula of the trigonometric relation between the sine and cosine function as:
${{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x$
Then, by substituting the value of ${{\cos }^{-1}}x$ found above in the given expression to solve the expression, we get:
$4{{\sin }^{-1}}x+\dfrac{\pi }{2}-{{\sin }^{-1}}x=\pi $
Then, by solving the above expression in the appropriate way as required, we get:
$\begin{align}
  & 3{{\sin }^{-1}}x=\pi -\dfrac{\pi }{2} \\
 & \Rightarrow 3{{\sin }^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\sin }^{-1}}x=\dfrac{\pi }{6} \\
\end{align}$
Now, by taking sine function on both sides to get the function only in x terms, we get:
$\sin \left( {{\sin }^{-1}}x \right)=\sin \left( \dfrac{\pi }{6} \right)$
Then, by solving the above expression, we get the value of x as:
$x=\dfrac{1}{2}$
So, we get the value of x as$\dfrac{1}{2}$.
Hence, option (d) is correct.

Note:
Now, to solve these type of the questions we need to know some of the basic conversions of trigonometry beforehand so that we can easily proceed in these type of questions. Then, some of the basic conversions of trigonometry are:
$\begin{align}
  & \sin \theta =\cos \left( {{90}^{\circ }}-\theta \right) \\
 & {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\
\end{align}$
Moreover, we must be careful while solving the different values of sine function as if we take any wrong value, we will get the wrong answer.