Question

If $ \cos x = \left( {\dfrac{1}{4}} \right) $ , where x lies in the fourth quadrant. How do you find the value of $ \sin \left( {x - \dfrac{\pi }{6}} \right) $ ?

Answer 1

Hint: In the given problem, we are required to calculate sine of a compound angle that is the sum of two angles; one whose cosine is given to us and other angle is $ \left( {\dfrac{\pi }{6}} \right) $ . Such problems require basic knowledge of trigonometric ratios and formulae. Besides this, knowledge of concepts of algebraic rules and properties is extremely essential to answer these questions correctly.

Complete step-by-step answer:
In the given problem, we are required to find the sine of a compound angle. So, we should remember the compound angle formula for sine $ \sin (A + B) = \sin A\cos B + \cos A\sin B $ .
Hence, using $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $ , we get,
 $ \Rightarrow {\sin ^2}x + {\left( {\dfrac{1}{4}} \right)^2} = 1 $
 $ \Rightarrow {\sin ^2}x = 1 - \dfrac{1}{{16}} $
 $ \Rightarrow {\sin ^2}x = \dfrac{{15}}{{16}} $
Now, we know that cosine function is positive in the fourth quadrant but sine function is negative in the fourth quadrant.
 $ \Rightarrow \sin x = - \sqrt {\dfrac{{15}}{{16}}} $
We know that the square root of $ 16 $ is $ 4 $ . So, we get,
 $ \Rightarrow \sin x = - \dfrac{{\sqrt {15} }}{4} $
So, the value of sine of angle x is $ - \left( {\dfrac{{\sqrt {15} }}{4}} \right) $ .
 $ \sin \left( {x - \dfrac{\pi }{6}} \right) = \sin x\cos \left( {\dfrac{\pi }{6}} \right) - \cos x\sin \left( {\dfrac{\pi }{6}} \right) $
 $ \Rightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \left( {\dfrac{{ - \sqrt {15} }}{4}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) - \left( {\dfrac{1}{4}} \right)\left( {\dfrac{1}{2}} \right) $
We know that $ \sqrt {15} = \sqrt 3 \times \sqrt 5 $ and $ \sqrt 3 \times \sqrt 3 = 3 $ . Hence, taking out $ 3 $ common from inside the square root, we get,
 $ \Rightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{{ - 3\sqrt 5 }}{8} - \dfrac{1}{8} $
 $ \Rightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = - \left( {\dfrac{{3\sqrt 5 + 1}}{8}} \right) $
So, the value of $ \sin \left( {x - \dfrac{\pi }{6}} \right) $ given $ \cos x = \left( {\dfrac{1}{4}} \right) $ where x lies in fourth quadrant is $ - \left( {\dfrac{{3\sqrt 5 + 1}}{8}} \right) $ .
So, the correct answer is “$ - \left( {\dfrac{{3\sqrt 5 + 1}}{8}} \right) $”.

Note: For finding the value of a trigonometric function for an angle given any other trigonometric ratio, we can use trigonometric identities. Then we find the required trigonometric ratio with help of basic trigonometric formulae and definitions of trigonometric ratios. Such questions require clarity of basic concepts of trigonometric functions as well as their inverse.