Question

What is ${{\sin }^{-1}}x+{{\sin }^{-1}}y$ ?

Answer 1

Hint : We first assume variables for the given terms ${{\sin }^{-1}}x$ and ${{\sin }^{-1}}y$ . We take trigonometric ratio of sine on both sides of $a+b=p$ . We use the formula of $\sin \left( a+b \right)=\sin a\cos b+\cos a\sin b$ . At the end we take inverse value to find the value of ${{\sin }^{-1}}x+{{\sin }^{-1}}y$ .

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

Note : We are only taking the positive value for the $\cos a=\sqrt{1-{{x}^{2}}}$ and $\cos b=\sqrt{1-{{y}^{2}}}$ . We use them using the formula of ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ .