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Question

Question: How do you simplify \[\cos \left( 2\arcsin x \right)\]?...

How do you simplify cos(2arcsinx)\cos \left( 2\arcsin x \right)?

Explanation

Solution

This type of question is based on the concept of trigonometry. We should first substitute α=arcsinx\alpha =\arcsin x. Arc of a trigonometric function is nothing but the inverse of that function. Thus, we get α=sin1(x)\alpha ={{\sin }^{-1}}\left( x \right). And the given function turns to cos(2α)\cos \left( 2\alpha \right). Now we have to simplify cos(2α)\cos \left( 2\alpha \right). Use the trigonometric identity cos2θ=12sin2θ\cos 2\theta =1-2{{\sin }^{2}}\theta and convert the assumed function to cos2α=12sin2α\cos 2\alpha =1-2{{\sin }^{2}}\alpha . We should then substitute α=sin1(x)\alpha ={{\sin }^{-1}}\left( x \right) in the obtained equation. Use the trigonometric identity sin(sin1θ)=θ\sin \left( {{\sin }^{-1}}\theta \right)=\theta in the RHS of the obtained equation and simplify the given function.

Complete step-by-step solution:
According to the question, we are asked to simplify cos(2arcsinx)\cos \left( 2\arcsin x \right).
We have been given the function is cos(2arcsinx)\cos \left( 2\arcsin x \right).
We know that arc means inverse of the function. ---------(1)
Therefore, the given function is cos(2sin1x)\cos \left( 2si{{n}^{-1}}x \right).
Now, let us assume α=sin1x\alpha =si{{n}^{-1}}x.
Therefore, the function (1) is cos2α\cos 2\alpha . ------(2)
We know that cos2θ=12sin2θ\cos 2\theta =1-2{{\sin }^{2}}\theta .
On using this trigonometric identity in equation (2), we get
cos2α=12sin2α\cos 2\alpha =1-2{{\sin }^{2}}\alpha
But we have assumed that α=sin1x\alpha =si{{n}^{-1}}x.
Substituting α\alpha in the above obtained equation, we get
cos(2sin1x)=12sin2(sin1x)\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\sin }^{2}}\left( si{{n}^{-1}}x \right)
On further simplification, we get
cos(2sin1x)=12(sin(sin1x))2\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\left( \sin \left( si{{n}^{-1}}x \right) \right)}^{2}}
We know that sin(sin1θ)=θ\sin \left( {{\sin }^{-1}}\theta \right)=\theta . Using this trigonometric identity of inverse in the above obtained equation, we get
cos(2sin1x)=12(x)2\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\left( x \right)}^{2}}
cos(2sin1x)=12x2\therefore \cos \left( 2si{{n}^{-1}}x \right)=1-2{{x}^{2}}
Hence, the simplified form of cos(2arcsinx)\cos \left( 2\arcsin x \right) is 12x21-2{{x}^{2}}.

Note: Don’t get confused by the term arc which means inverse of that trigonometric function. We should not make calculation mistakes based on sign conventions. Be thorough with the trigonometric identities to simplify this type of problems. It is advisable to first convert the given function to a simpler form and then solve. The final answer should be in terms of x only and not in terms of any other terms like α\alpha .