Question

The value of cos $ \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ is
A. $ \dfrac{3}{4} $
B. $ - \dfrac{3}{4} $
C. $ \dfrac{1}{16} $
D. $ \dfrac{1}{4} $

Answer 1

Hint: In this question, to find the value of cos $ \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ , we need to assume $ \left( {{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ = $ \theta $ . Then we will put the value of $ \left( {{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ = $ \theta $ in cos $ \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ , we get cos $ \left( {\dfrac{1}{2}\theta } \right) $ . Then we will try to find the value of cos $ \left( {\dfrac{1}{2}\theta } \right) $ .

Complete step-by-step answer:
We have cos $ \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $
Let $ \left( {{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ = $ \theta $
Putting the value of $ \left( {{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ in cos $ \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ .
We get: cos $ \left( {\dfrac{1}{2}\theta } \right) $
Here, we need to find the value of cos $ \left( {\dfrac{1}{2}\theta } \right) $ .
Now we have;
 $ \left( {{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ = $ \theta $
Taking $ {\cos ^{ - 1}} $ to RHS.
Now, $ \cos \theta = \dfrac{1}{8} $
 $ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} - 1 = \dfrac{1}{8} $ (Using the formula $ \cos 2\theta = 2{\cos ^2}\theta - 1 $ )
Taking -1 to RHS.
 $ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = 1 + \dfrac{1}{8} $
Taking LCM of ( $ 1 + \dfrac{1}{8} $ ), we get;
 $ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = \dfrac{9}{8} $
 $ \Rightarrow {\cos ^2}\dfrac{\theta }{2} = \dfrac{9}{{16}} $
 $ \Rightarrow \cos \dfrac{\theta }{2} = \dfrac{3}{4} $ (Ans.)
Thus, the value of cos $ \left( {\dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{1}{8}} \right) $ is $ \dfrac{3}{4} $ .

So, the correct answer is “Option A”.

Note: Here we have used double angle formula:
 $ \cos 2\theta = 2{\cos ^2}\theta - 1 $ = $ {\cos ^2}\theta - {\sin ^2}\theta $ = $ 1 - 2{\sin ^2}\theta $
 $ \sin 2\theta = 2\sin \theta \cos \theta $
 $ \tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} $
Inverse trigonometric functions: Inverse trigonometric functions are the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. Inverse trigonometric functions are also called “Arcus Functions”, “Anti-trigonometric functions” or “cyclometric functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent.