Question

What is the value of $\cos \left( 2{{\sin }^{-1}}x \right)?$

Answer 1

Hint: We will use the known trigonometric identities to solve the given problem. We will rearrange the given function by substitution. And then, we will use the trigonometric identity given by ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}.$

Complete step by step solution:
Let us consider the given trigonometric function $\cos \left( 2{{\sin }^{-1}}x \right).$
To find the simplified form of the given function, we need to rearrange the given function by substitution.
Let us equate the term inside the bracket which is an inverse trigonometric function with a parameter.
Let us suppose that ${{\sin }^{-1}}x=\theta .$
Now, from this, we will get $x=\sin \theta $
Now, let us substitute for the inverse Sine function in the given function.
As a result, we will get $\cos 2\theta .$
Let us recall the trigonometric identity given by ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}.$
We are going to rearrange this identity by transposing $2$ from the right-hand side to the left-hand side.
We will get $2{{\sin }^{2}}x=1-\cos 2x.$
Now, we are going to transpose $\cos 2x$ from the right-hand side to the left-hand side and $2{{\sin }^{2}}x$ from the left-hand side to the right-hand side of the equation.
Now, as a result of transposing the mentioned terms accordingly, we will get $\cos 2x=1-2{{\sin }^{2}}x.$
Now, using the above identity, we can write the given function as $\cos 2\theta =1-2{{\sin }^{2}}\theta .$
And since $\sin \theta =x,$ we can write the above equation as $\cos 2\theta =1-2{{x}^{2}}.$
And since we have $\theta ={{\sin }^{2}}x,$ we will get $\cos \left( 2{{\sin }^{-1}}x \right)=1-{2{x}^{2}}.$
Hence the answer is $\cos \left( 2{{\sin }^{-1}}x \right)=1-{2{x}^{2}}.$

Note: We should always remember the trigonometric identity given by ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}.$ This can be rearranged, as we know, as $2\cos 2x=1+2{{\cos }^{2}}x.$ And now, we can transpose the terms accordingly to get $2{{\cos }^{2}}x=2\cos 2x-1.$