
How do you simplify \[\cos \left( 2\arcsin x \right)\]?
Answer
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Hint: This type of question is based on the concept of trigonometry. We should first substitute \[\alpha =\arcsin x\]. Arc of a trigonometric function is nothing but the inverse of that function. Thus, we get \[\alpha ={{\sin }^{-1}}\left( x \right)\]. And the given function turns to \[\cos \left( 2\alpha \right)\]. Now we have to simplify \[\cos \left( 2\alpha \right)\]. Use the trigonometric identity \[\cos 2\theta =1-2{{\sin }^{2}}\theta \] and convert the assumed function to \[\cos 2\alpha =1-2{{\sin }^{2}}\alpha \]. We should then substitute \[\alpha ={{\sin }^{-1}}\left( x \right)\] in the obtained equation. Use the trigonometric identity \[\sin \left( {{\sin }^{-1}}\theta \right)=\theta \] 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 \[\cos \left( 2\arcsin x \right)\].
We have been given the function is \[\cos \left( 2\arcsin x \right)\].
We know that arc means inverse of the function. ---------(1)
Therefore, the given function is \[\cos \left( 2si{{n}^{-1}}x \right)\].
Now, let us assume \[\alpha =si{{n}^{-1}}x\].
Therefore, the function (1) is \[\cos 2\alpha \]. ------(2)
We know that \[\cos 2\theta =1-2{{\sin }^{2}}\theta \].
On using this trigonometric identity in equation (2), we get
\[\cos 2\alpha =1-2{{\sin }^{2}}\alpha \]
But we have assumed that \[\alpha =si{{n}^{-1}}x\].
Substituting \[\alpha \] in the above obtained equation, we get
\[\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\sin }^{2}}\left( si{{n}^{-1}}x \right)\]
On further simplification, we get
\[\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\left( \sin \left( si{{n}^{-1}}x \right) \right)}^{2}}\]
We know that \[\sin \left( {{\sin }^{-1}}\theta \right)=\theta \]. Using this trigonometric identity of inverse in the above obtained equation, we get
\[\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\left( x \right)}^{2}}\]
\[\therefore \cos \left( 2si{{n}^{-1}}x \right)=1-2{{x}^{2}}\]
Hence, the simplified form of \[\cos \left( 2\arcsin x \right)\] is \[1-2{{x}^{2}}\].
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 \[\alpha \].
Complete step-by-step solution:
According to the question, we are asked to simplify \[\cos \left( 2\arcsin x \right)\].
We have been given the function is \[\cos \left( 2\arcsin x \right)\].
We know that arc means inverse of the function. ---------(1)
Therefore, the given function is \[\cos \left( 2si{{n}^{-1}}x \right)\].
Now, let us assume \[\alpha =si{{n}^{-1}}x\].
Therefore, the function (1) is \[\cos 2\alpha \]. ------(2)
We know that \[\cos 2\theta =1-2{{\sin }^{2}}\theta \].
On using this trigonometric identity in equation (2), we get
\[\cos 2\alpha =1-2{{\sin }^{2}}\alpha \]
But we have assumed that \[\alpha =si{{n}^{-1}}x\].
Substituting \[\alpha \] in the above obtained equation, we get
\[\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\sin }^{2}}\left( si{{n}^{-1}}x \right)\]
On further simplification, we get
\[\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\left( \sin \left( si{{n}^{-1}}x \right) \right)}^{2}}\]
We know that \[\sin \left( {{\sin }^{-1}}\theta \right)=\theta \]. Using this trigonometric identity of inverse in the above obtained equation, we get
\[\Rightarrow \cos \left( 2si{{n}^{-1}}x \right)=1-2{{\left( x \right)}^{2}}\]
\[\therefore \cos \left( 2si{{n}^{-1}}x \right)=1-2{{x}^{2}}\]
Hence, the simplified form of \[\cos \left( 2\arcsin x \right)\] is \[1-2{{x}^{2}}\].
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 \[\alpha \].
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