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Question: Using trigonometric identities, find the value of \(\cos \left( 2{{\sin }^{-1}}\dfrac{1}{3} \right)\...

Using trigonometric identities, find the value of cos⁡(2sin⁡−113)\cos \left( 2{{\sin }^{-1}}\dfrac{1}{3} \right)

Explanation

Solution

Hint: In this question, we should note that we are asked to find the cosine of the given expression involving sine inverse. Therefore, we should use some trigonometric identities to convert the cosine function in terms of sine so that the sine and cosine functions may get cancelled and from it we can obtain the required answer to the given question.

Complete step-by-step answer:

The expression we are asked to evaluate is cos⁡(2sin⁡−113)\cos \left( 2{{\sin }^{-1}}\dfrac{1}{3} \right)……………………….(1.1)
Now, we know that the formula for the cosine of twice angle is given by
cos⁡(2θ)=1−2sin⁡2(θ)..................(1.2)\cos \left( 2\theta \right)=1-2{{\sin }^{2}}\left( \theta \right)..................(1.2)
Now, if we take θ=sin⁡−1(13)\theta ={{\sin }^{-1}}\left( \dfrac{1}{3} \right) in equation (1.2), we obtain
cos⁡(2sin⁡−113)=1−2sin⁡2(sin⁡−113) =1−2(sin⁡(sin⁡−113))2..................(1.3) \begin{aligned} & \cos \left( 2{{\sin }^{-1}}\dfrac{1}{3} \right)=1-2{{\sin }^{2}}\left( {{\sin }^{-1}}\dfrac{1}{3} \right) \\\ & =1-2{{\left( \sin \left( {{\sin }^{-1}}\dfrac{1}{3} \right) \right)}^{2}}..................(1.3) \\\ \end{aligned}
Now, as sin⁡−1{{\sin }^{-1}} is the inverse function of sin, we should have
sin⁡(sin⁡−1x)=x\sin \left( {{\sin }^{-1}}x \right)=x for any x in its domain.
Using this in equation (1.3) with x=13x=\dfrac{1}{3}, we get
cos⁡(2sin⁡−113)=1−2(sin⁡(sin⁡−113))2=1−2×(13)2 =1−2×19=9−29=79 \begin{aligned} & \cos \left( 2{{\sin }^{-1}}\dfrac{1}{3} \right)=1-2{{\left( \sin \left( {{\sin }^{-1}}\dfrac{1}{3} \right) \right)}^{2}}=1-2\times {{\left( \dfrac{1}{3} \right)}^{2}} \\\ & =1-2\times \dfrac{1}{9}=\dfrac{9-2}{9}=\dfrac{7}{9} \\\ \end{aligned}
Thus, we obtain
cos⁡(2sin⁡−113)=79\cos \left( 2{{\sin }^{-1}}\dfrac{1}{3} \right)=\dfrac{7}{9}
Which is the required answer to this question.

Note: In this question, we converted the given expression in terms of sin to cancel out sin⁡−1{{\sin }^{-1}}. However, we could also have explicitly found out the value of sin⁡−1(13){{\sin }^{-1}}\left( \dfrac{1}{3} \right) and then found out the cosine of the resulting term in the parenthesis. However, as 13\dfrac{1}{3} does not correspond to the sine of any standard angle, the angle would come out as an approximate value, therefore, the calculations by explicitly obtaining the value of the term in parenthesis would be difficult and also not perfectly accurate.

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