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Question: How do you evaluate \(\cos \left( {2{{\cos }^{ - 1}}\left( {\dfrac{1}{7}} \right)} \right)\)?...

How do you evaluate cos⁡(2cos⁡−1(17))\cos \left( {2{{\cos }^{ - 1}}\left( {\dfrac{1}{7}} \right)} \right)?

Explanation

Solution

Here we have to find the value of the given trigonometric function cos⁡(2cos⁡−1(17))\cos \left( {2{{\cos }^{ - 1}}\left( {\dfrac{1}{7}} \right)} \right). We will use basic trigonometric rules and identities such as cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1 and if cos⁡−1(x)=y{\cos ^{ - 1}}(x) = y then cos⁡y=x\cos y = x. In order to solve this question we first assume that cos⁡−1(17)=θ{\cos ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \theta then we will proceed to get the required result.

Complete step by step solution:
The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sin⁡,  cos⁡,  tan⁡,  csc⁡,  sec⁡, cot⁡ \sin ,\,\,\cos ,\,\,\tan ,\,\,\csc ,\,\,\sec ,\,\cot \,. The conventional symbol to represent the inverse trigonometric functions is using arc- prefix like arcsin⁡(x),arccos⁡(x)\arcsin \left( x \right),\arccos \left( x \right) or we can represent as sin⁡−1x,  cos⁡−1x,  tan⁡−1x{\sin ^{ - 1}}x,\,\,{\cos ^{ - 1}}x,\,\,{\tan ^{ - 1}}x. sin⁡−1x,  cos⁡−1x,  tan⁡−1x{\sin ^{ - 1}}x,\,\,{\cos ^{ - 1}}x,\,\,{\tan ^{ - 1}}x denote angles or real numbers whose sin⁡\sin is xx, cos⁡\cos is xx and tan⁡\tan is xx provided that the results given are numerically smallest as possible. Where sin⁡−1x,  tan⁡−1x,   sec⁡−1x{\sin ^{ - 1}}x,\,\,{\tan ^{ - 1}}x,\,\,\,{\sec ^{ - 1}}x are increasing functions and cos⁡−1x  ,  cot⁡−1x,    csc⁡−1x{\cos ^{ - 1}}x\,\,,\,\,{\cot ^{ - 1}}x,\,\,\,\,{\csc ^{ - 1}}x are decreasing functions.
Here, we have to evaluate the value of the given trigonometric function cos⁡(2cos⁡−1(17))\cos \left( {2{{\cos }^{ - 1}}\left( {\dfrac{1}{7}} \right)} \right).
Let cos⁡−1(17)=θ{\cos ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \theta
Then, cos⁡θ=17\cos \theta = \dfrac{1}{7}
Substituting cos⁡−1(17)=θ{\cos ^{ - 1}}\left( {\dfrac{1}{7}} \right) = \theta in the given function. we get,
⇒cos⁡(2θ)\Rightarrow \cos \left( {2\theta } \right)
We know that cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1
So, we have
⇒cos⁡2θ=2cos⁡2θ−1\Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1
Putting cos⁡θ=17\cos \theta = \dfrac{1}{7} in the above equation. We get,
⇒2cos⁡2θ−1=2(17)2−1\Rightarrow 2{\cos ^2}\theta - 1 = 2{\left( {\dfrac{1}{7}} \right)^2} - 1
⇒2×149−1\Rightarrow 2 \times \dfrac{1}{{49}} - 1
Solving the above equation. We get,
⇒249−1\Rightarrow \dfrac{2}{{49}} - 1
Simplifying the above equation by taking L.C.M (Least Common Factor). We get,
⇒2−4949\Rightarrow \dfrac{{2 - 49}}{{49}}
⇒−4749\Rightarrow \dfrac{{ - 47}}{{49}}
Hence the value of the function cos⁡(2cos⁡−1(17))\cos \left( {2{{\cos }^{ - 1}}\left( {\dfrac{1}{7}} \right)} \right) is −4749\dfrac{{ - 47}}{{49}}.

Note:
In order to solve these types of problems we must know all the inverse trigonometric formulas and basic trigonometric identities. Inverse trigonometric functions are also known as anti- trigonometric functions, arcus functions and cyclometric functions. These inverse trigonometric functions help us to find out any angles with any of the trigonometric ratios and derived from the properties of trigonometric functions. These types of problems can be solved by the trigonometric formulas such as cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2{\sin ^2}x . Likewise if the function is in sin⁡x\sin x we can use the identities of sin⁡x\sin x.