Question

How do you find $\cos \left( {{\sin }^{-1}}x-{{\cos }^{-1}}y \right)$?

Answer 1

We first assume variables for the given terms ${{\sin }^{-1}}x$ and ${{\cos }^{-1}}y$ as a and b respectively. We take the trigonometric ratio of cos on both sides of $a-b$. We find both ratio value of cos and sin for the angles a and b. then we use the formula of $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$. At the end we put the values to find the solution.

Complete step by step solution:
Let ${{\sin }^{-1}}x=a$ and ${{\cos }^{-1}}y=b$. From the inverse law we get $\sin a=x$ and $\cos b=y$.
Therefore, we need to find the value of ${{\sin }^{-1}}x-{{\cos }^{-1}}y=a-b$.
We need to find the value of $\cos \left( {{\sin }^{-1}}x-{{\cos }^{-1}}y \right)$ equal to $\cos \left( a-b \right)$.
We take the trigonometric ratio of cos on both sides of $a-b$.
We now use the theorem of $\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b$.
From $\sin a=x$ and $\cos b=y$, we find the values of $\cos a$ and $\sin b$.
So, we get $\cos a=\sqrt{1-{{\sin }^{2}}a}=\sqrt{1-{{x}^{2}}}$ and $\sin b=\sqrt{1-{{\cos }^{2}}b}=\sqrt{1-{{y}^{2}}}$.
We place the values and get
$\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b=y\sqrt{1-{{x}^{2}}}+x\sqrt{1-{{y}^{2}}}$.
Therefore, we get $\cos \left( {{\sin }^{-1}}x-{{\cos }^{-1}}y \right)=y\sqrt{1-{{x}^{2}}}+x\sqrt{1-{{y}^{2}}}$.

Note:
We are only taking the positive value for the $\cos a=\sqrt{1-{{x}^{2}}}$ and $\sin b=\sqrt{1-{{y}^{2}}}$. We use them using the formula of ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$. Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$.