Question

How do you simplify $ \cos \left( 2{{\sin }^{-1}}x \right) $ ?

Answer 1

Hint: To simplify the above trigonometric expression i.e. $ \cos \left( 2{{\sin }^{-1}}x \right) $ . We are going to assume $ {{\sin }^{-1}}x $ as $ \theta $ and then substitute $ \theta $ in place of $ {{\sin }^{-1}}x $ in the above expression. Now, we have assumed that $ {{\sin }^{-1}}x=\theta $ so we are taking sine on both the sides then we get $ \sin \left( {{\sin }^{-1}}x \right)=\sin \theta $ . After that we are going to use the property that $ \sin \left( {{\sin }^{-1}} \right)=1 $ . Then we get $ x=\sin \theta $ . Now, when we have substituted $ \theta $ in place of $ {{\sin }^{-1}}x $ then the expression will look like $ \cos \left( 2\theta \right) $ . Then we will use the property of $ \cos 2\theta =1-2{{\sin }^{2}}\theta $ and then we will use $ x=\sin \theta $ to further simplify it.

Complete step by step answer:
The trigonometric expression which we have to simplify is as follows:
$\Rightarrow$ $ \cos \left( 2{{\sin }^{-1}}x \right) $
In the above expression, we are assuming $ {{\sin }^{-1}}x=\theta $ and then replace $ {{\sin }^{-1}}x $ by $ \theta $ and we get,
$\Rightarrow$ $ \cos \left( 2\theta \right) $
Now, taking sine on both the sides of $ {{\sin }^{-1}}x=\theta $ we get,
 $ \sin \left( {{\sin }^{-1}}x \right)=\sin \theta $
We know that multiplying a number or expression by its inverse will give 1 and the above equation will look like:
 $ x=\sin \theta $
We know the trigonometric property that:
 $ \cos 2\theta =1-2{{\sin }^{2}}\theta $
Using the above relation in $ \cos \left( 2\theta \right) $ we get,
 $ \cos 2\theta =1-2{{\sin }^{2}}\theta $
Now, substituting x in place of $ \sin \theta $ in the above equation we get,
$\Rightarrow$ $ 1-2{{x}^{2}} $
Hence, we have simplified the above trigonometric expression to $ 1-2{{x}^{2}} $ .

Note:
The alternate approach to solve the above problem by assuming $ 2{{\sin }^{-1}}x=\theta $ then taking sine on both the sides and the above equation will look like:
 $ \sin \left( 2{{\sin }^{-1}}x \right)=\sin \theta $
In the above, let us take $ {{\sin }^{-1}}x=\phi $ in the above we get,
 $ \sin \left( 2\phi \right)=\sin \theta $
We are going to use the identity $ \sin 2\phi =2\sin \phi \cos \phi $ in the above and we get,
 $ 2\sin \phi \cos \phi =\sin \theta $
Taking sine on both the sides in $ {{\sin }^{-1}}x=\phi $ we get,
 $ \begin{align}
  & \sin \left( {{\sin }^{-1}}x \right)=\sin \phi \\
 & \Rightarrow x=\sin \phi \\
\end{align} $
We also know the identity that $ \cos \phi =\sqrt{1-{{\sin }^{2}}\phi } $ so substituting $ \sin \phi =x $ in this property we get,
 $ \cos \phi =\sqrt{1-{{x}^{2}}} $
Now, substituting the value of $ \sin \phi \And \cos \phi $ that we have solved above in $ 2\sin \phi \cos \phi =\sin \theta $ we get,
 $ \sin \theta =2x\sqrt{1-{{x}^{2}}} $
Now, replacing $ 2{{\sin }^{-1}}x=\theta $ in $ \cos \left( 2{{\sin }^{-1}}x \right) $ we get,
$\Rightarrow$ $ \cos \theta =\sqrt{1-{{\sin }^{2}}\theta } $
Using $ \sin \theta =2x\sqrt{1-{{x}^{2}}} $ in the above equation we get,
 $ \begin{align}
  & \cos \theta =\sqrt{1-{{\left( 2x\sqrt{1-{{x}^{2}}} \right)}^{2}}} \\
 & \Rightarrow \cos \theta =\sqrt{1-\left( 4{{x}^{2}}\left( 1-{{x}^{2}} \right) \right)} \\
 & \Rightarrow \cos \theta =\sqrt{1-4{{x}^{2}}+4{{x}^{4}}} \\
 & \Rightarrow \cos \theta =\sqrt{{{\left( 1-2{{x}^{2}} \right)}^{2}}} \\
 & \Rightarrow \cos \theta =1-2{{x}^{2}} \\
\end{align} $