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Question: Find the value of \[\dfrac{{{{\cos }^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right)}}{{{{\sin }^{ - 1}}\...

Find the value of cos⁡−1(4149)sin⁡−1(27)\dfrac{{{{\cos }^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right)}}{{{{\sin }^{ - 1}}\left( {\dfrac{2}{7}} \right)}}
A. 4
B. 3
C. 2
D. 1

Explanation

Solution

In this question, we need to determine the value of the given trigonometric equation. For this, we will use the trigonometric identities along with the inverse trigonometric function.
Trigonometric functions used in this question:
cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta
cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x

Complete step by step solution:
We have to determine the value of the trigonometric function cos⁡−1(4149)sin⁡−1(27)−−(i)\dfrac{{{{\cos }^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right)}}{{{{\sin }^{ - 1}}\left( {\dfrac{2}{7}} \right)}} - - (i)
Let us expand the numerator of the given trigonometric equation, so this can be written as
cos⁡−1(4149)=cos⁡−1(1−849){\cos ^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right) = {\cos ^{ - 1}}\left( {1 - \dfrac{8}{{49}}} \right)
Now, write 8 as the product of 2 and 4 in the above equation, we get

cos⁡−1(4149)=cos⁡−1(1−2×2272) =cos⁡−1(1−2(27)2)−−(ii)  {\cos ^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right) = {\cos ^{ - 1}}\left( {1 - \dfrac{{2 \times {2^2}}}{{{7^2}}}} \right) \\\ = {\cos ^{ - 1}}\left( {1 - 2{{\left( {\dfrac{2}{7}} \right)}^2}} \right) - - (ii) \\\

Let us equate the denominator of the given trigonometric equation equal to θ\theta ; hence we can say

sin⁡−1(27)=θ ⇒sin⁡θ=27  {\sin ^{ - 1}}\left( {\dfrac{2}{7}} \right) = \theta \\\ \Rightarrow \sin \theta = \dfrac{2}{7} \\\

Since the value of sin⁡θ=27\sin \theta = \dfrac{2}{7}, hence we can write equation (ii) as
cos⁡−1(4149)=cos⁡−1(1−2sin⁡2θ)−−(iii){\cos ^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right) = {\cos ^{ - 1}}\left( {1 - 2{{\sin }^2}\theta } \right) - - (iii)
Now, as we know the trigonometric identity cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta , hence we can write equation (iii) as
cos⁡−1(4149)=cos⁡−1(cos⁡2θ)−−−−(iv){\cos ^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right) = {\cos ^{ - 1}}\left( {\cos 2\theta } \right) - - - - (iv)
Now, using the inverse trigonometric identity cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x, equation (iv) can be simplified as:
cos⁡−1(4149)=2θ−−−−(v){\cos ^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right) = 2\theta - - - - (v)
Substituting the values from equation (v) and sin⁡−1(27)=θ{\sin ^{ - 1}}\left( {\dfrac{2}{7}} \right) = \theta in the equation (i), we get
cos⁡−1(4149)sin⁡−1(27)=2θθ=2\dfrac{{{{\cos }^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right)}}{{{{\sin }^{ - 1}}\left( {\dfrac{2}{7}} \right)}} = \dfrac{{2\theta }}{\theta } = 2
Hence, cos⁡−1(4149)sin⁡−1(27)=2\dfrac{{{{\cos }^{ - 1}}\left( {\dfrac{{41}}{{49}}} \right)}}{{{{\sin }^{ - 1}}\left( {\dfrac{2}{7}} \right)}} = 2

Therefore option (C) is correct

Note:
It is worth noting down here that we have simplified the term inside the numerator such that the equivalent absolute value can be seen similar to the term available in the denominator. This is done so as to use the inverse trigonometric identity of cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2{\sin ^2}\theta and then, cos⁡−1(cos⁡x)=x{\cos ^{ - 1}}\left( {\cos x} \right) = x