Question

How do you prove the given identity ${{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)=\pi $?

Answer 1

Hint: We start solving the problem by considering the L.H.S (Left Hand Side) of the given identity. We then apply the cosine function on both sides of the assumed L.H.S and make use of the fact that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ to proceed through the problem. We then make use of the fact that $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ and then make the necessary calculations to complete the proof of required identity.

Complete step by step answer:
According to the problem, we are asked to prove the given identity ${{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)=\pi $, $-1\le x\le 1$.
Let us consider L.H.S (Left Hand Side) of the given identity and try to prove it equal to the given R.H.S (Right Hand Side).
So, let us consider ${{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)$.
Let us assume ${{\cos }^{-1}}\left( -x \right)=A\Leftrightarrow \cos A=-x$ and ${{\cos }^{-1}}\left( x \right)=B\Leftrightarrow \cos B=x$.
So, we have ${{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)=A+B$ ---(1).
Now, let us apply cosine on both sides of the equation (1).
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right) \right)=\cos \left( A+B \right)$ ---(2).
We know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. Let us use this result in equation (2).
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right) \right)=\cos A\cos B-\sin A\sin B$ ---(3).
We know that $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$. So, we have $\sin A=\sqrt{1-{{\left( -x \right)}^{2}}}=\sqrt{1-{{x}^{2}}}$ and $\sin B=\sqrt{1-{{x}^{2}}}$. Let us use these results in equation (3).
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right) \right)=\left( \left( -x \right)\times x \right)-\left( \left( \sqrt{1-{{x}^{2}}} \right)\times \left( \sqrt{1-{{x}^{2}}} \right) \right)$.
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right) \right)=\left( -{{x}^{2}} \right)-\left( 1-{{x}^{2}} \right)$.
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right) \right)=-{{x}^{2}}-1+{{x}^{2}}$.
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right) \right)=-1$.
$\Rightarrow {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)={{\cos }^{-1}}\left( -1 \right)$ ---(4).
We know that the principal solution of ${{\cos }^{-1}}\left( -1 \right)=\pi $. Let us use this result in equation (4).
$\Rightarrow {{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)=\pi $.
We can see that the considered L.H.S (Left Hand Side) of the identity is equal to the given R.H.S (Right Hand Side) of the identity.
$\therefore $ We have proved the given identity ${{\cos }^{-1}}\left( -x \right)+{{\cos }^{-1}}\left( x \right)=\pi $.

Note:
We should perform each step carefully in order to avoid confusion and calculation mistakes. Here we have assumed that the values of x lie in the interval $-1\le x\le 1$ to prove the given identity otherwise, the identity doesn’t hold true. We can also solve this problem by applying secant to the L.H.S (Left Hand Side) of the identity. Similarly, we can expect problems to prove the result ${{\sin }^{-1}}\left( x \right)+{{\sin }^{-1}}\left( -x \right)=0$.