Question

Find the value x:
${{\sin }^{-1}}x+{{\sin }^{-1}}\left( 1-x \right)={{\cos }^{-1}}x$.

Answer 1

Hint: First of all take cos on both the sides of the given equation then in the left hand side of the equation, use the formula cos (A + B) =cos A cos B – sin A sin B. After that we will find the equation is in variable x then solve for x.

Complete step-by-step answer:
We are taking cos on both the sides in the expression given in the question as follows:
${{\sin }^{-1}}x+{{\sin }^{-1}}\left( 1-x \right)={{\cos }^{-1}}x$
$\cos \left( {{\sin }^{-1}}x+{{\sin }^{-1}}\left( 1-x \right) \right)=\cos \left( {{\cos }^{-1}}x \right)$
In the above expression, consider ${{\sin }^{-1}}x$ as A and ${{\sin }^{-1}}\left( 1-x \right)$ as B then the left hand side of the expression looks like cos (A+ B) which is equal to cos A cos B – sin A sin B and $\cos \left( {{\cos }^{-1}}x \right)$ is reduced to x. Now, substituting these conditions in the above expression we get,
$\cos ({{\sin }^{-1}}x)\cos \left( {{\sin }^{-1}}\left( 1-x \right) \right)-\sin \left( {{\sin }^{-1}}x \right)\sin \left( {{\sin }^{-1}}\left( 1-x \right) \right)=x$…………….Eq. (1)
Now, let us assume ${{\sin }^{-1}}x=\theta $. This ${{\sin }^{-1}}x$ is the one which is the angle of $\cos \left( {{\sin }^{-1}}x \right)$ in the above expression.
Solving the value of $\sin \left( {{\sin }^{-1}}x \right)$ as follows:
We have assumed that:
${{\sin }^{-1}}x=\theta $.
Now taking sin on both the sides we get,
$\begin{align}
  & \sin \left( {{\sin }^{-1}}x \right)=\sin \theta \\
 & \Rightarrow x=\sin \theta \\
\end{align}$
In the above steps, we have shown that $\sin \left( {{\sin }^{-1}}x \right)$ equals x. This is due to the fact that when we write any number, polynomial or any expression with its inverse then it reduces to 1 or identity. Also, we have shown that sin θ = x so $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$ and cos θ in terms of x will be $\sqrt{1-{{x}^{2}}}$.
So, we can write $\cos \left( {{\sin }^{-1}}x \right)=\sqrt{1-{{x}^{2}}}$. Similarly, we can find the value of -$\cos \left( {{\sin }^{-1}}\left( 1-x \right) \right)\text{, }\sin \left( {{\sin }^{-1}}\left( 1-x \right) \right)$.
In $\cos \left( {{\sin }^{-1}}\left( 1-x \right) \right)$, let us assume that ${{\sin }^{-1}}\left( 1-x \right)=\alpha $. Now, taking sin on both the sides we get,
$1-x=\sin \alpha $
We can write $\cos \left( {{\sin }^{-1}}\left( 1-x \right) \right)$ as cos α so the value of cos α in terms of x is \[\sqrt{1-\left( 1-{{x}^{2}} \right)}\].
Hence, $\cos \left( {{\sin }^{-1}}\left( 1-x \right) \right)=\sqrt{1-{{\left( 1-x \right)}^{2}}}$
Now, $\sin \left( {{\sin }^{-1}}\left( 1-x \right) \right)$is equal to 1 – x. Similarly, we can also say that $\sin \left( {{\sin }^{-1}}x \right)=x$.
Substituting $\cos \left( {{\sin }^{-1}}x \right),\cos \left( {{\sin }^{-1}}\left( 1-x \right) \right)\text{,sin}\left( {{\sin }^{-1}}x \right),\sin \left( {{\sin }^{-1}}\left( 1-x \right) \right)$in Eq. (1) we get,
$\begin{align}
  & \sqrt{1-{{x}^{2}}}\sqrt{1-{{\left( 1-x \right)}^{2}}}-x\left( 1-x \right)=x \\
 & \Rightarrow \sqrt{1-{{x}^{2}}}\sqrt{1-{{\left( 1-x \right)}^{2}}}-x+{{x}^{2}}=x \\
 & \Rightarrow \sqrt{1-{{x}^{2}}}\sqrt{1-{{\left( 1-x \right)}^{2}}}=2x-{{x}^{2}} \\
 & \\
\end{align}$
On squaring both the sides we get,
$\begin{align}
  & \left( 1-{{x}^{2}} \right)\left( 1-{{\left( 1-x \right)}^{2}} \right)={{\left( 2x-{{x}^{2}} \right)}^{2}} \\
 & \Rightarrow \left( 1-{{x}^{2}} \right)\left( 1-\left( 1+{{x}^{2}}-2x \right) \right)=4{{x}^{2}}+{{x}^{4}}-4{{x}^{3}} \\
 & \Rightarrow \left( 1-{{x}^{2}} \right)\left( 1-1-{{x}^{2}}+2x \right)=4{{x}^{2}}+{{x}^{4}}-4{{x}^{3}} \\
 & \Rightarrow \left( 1-{{x}^{2}} \right)\left( -{{x}^{2}}+2x \right)=4{{x}^{2}}+{{x}^{4}}-4{{x}^{3}} \\
 & \Rightarrow -{{x}^{2}}+2x+{{x}^{4}}-2{{x}^{3}}=4{{x}^{2}}+{{x}^{4}}-4{{x}^{3}} \\
\end{align}$
In the above step ${x}^{4}$ will be cancelled out from both the sides and then the above expression will look like:
$\begin{align}
  & 4{{x}^{3}}-2{{x}^{3}}+2x-5{{x}^{2}}=0 \\
 & \Rightarrow 2{{x}^{3}}-5{{x}^{2}}+2x=0 \\
 & \Rightarrow x\left( 2{{x}^{2}}-5x+2 \right)=0 \\
 & \\
\end{align}$
The solutions of the above equation are as follows:
$x=0\text{, 2}{{x}^{2}}-5x+2=0$
Solving the quadratic equation we get,
$\begin{align}
  & 2{{x}^{2}}-5x+2=0 \\
 & \Rightarrow 2{{x}^{2}}-4x-x+2=0 \\
 & \Rightarrow 2x\left( x-2 \right)-1\left( x-2 \right)=0 \\
 & \Rightarrow \left( 2x-1 \right)\left( x-2 \right)=0 \\
 & x=\dfrac{1}{2},x=2 \\
\end{align}$
Here, we have got 3 solutions $x=0,\dfrac{1}{2},2$.
Now, 2 cannot be a solution because ${{\sin }^{-1}}\left( 2 \right)\And {{\cos }^{-1}}\left( 2 \right)$ are not defined.
Hence, the solutions are $x=0,\dfrac{1}{2}$.

Note: Always check the solutions that you got from solving the equation by putting the solutions one by one in the mother equation. Like in the above question, we are getting 3 solutions, 1 is rejected because ${{\sin }^{-1}}\left( 2 \right)\And {{\cos }^{-1}}\left( 2 \right)$ are not defined. Now, we are going to check the other solutions by putting these solutions in ${{\sin }^{-1}}x+{{\sin }^{-1}}\left( 1-x \right)={{\cos }^{-1}}x$.
Putting x = 0 in the above equation we get,
$\begin{align}
  & {{\sin }^{-1}}0+{{\sin }^{-1}}\left( 1-0 \right)={{\cos }^{-1}}0 \\
 & 0+\dfrac{\pi }{2}=\dfrac{\pi }{2} \\
\end{align}$
Hence, x = 0 is satisfying the given equation. Now, putting $x=\dfrac{1}{2}$ in the given equation, we get: $\begin{align}
  & {{\sin }^{-1}}\dfrac{1}{2}+{{\sin }^{-1}}\left( 1-\dfrac{1}{2} \right)={{\cos }^{-1}}\dfrac{1}{2} \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{1}{2}+{{\sin }^{-1}}\dfrac{1}{2}={{\cos }^{-1}}\dfrac{1}{2} \\
 & \Rightarrow 2{{\sin }^{-1}}\dfrac{1}{2}={{\cos }^{-1}}\dfrac{1}{2} \\
 & \Rightarrow 2\dfrac{\pi }{6}=\dfrac{\pi }{3} \\
 & \Rightarrow \dfrac{\pi }{3}=\dfrac{\pi }{3} \\
\end{align}$
Hence, $x=\dfrac{1}{2}$ is satisfying the given equation.