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 $ .