Question

${{\cos }^{2}}\left( \dfrac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \dfrac{\pi }{6}-\theta \right)$ is equal to
A. $\dfrac{1}{2}\cos 2\theta $
B. $-\dfrac{1}{2}\cos 2\theta $
C. 0
D. 1

Answer 1

Hint: We are given the question ${{\cos }^{2}}\left( \dfrac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \dfrac{\pi }{6}-\theta \right)$ and we have to choose the option which is equal to our given question. As our question is in the form of square of cos and sin so we solve the equation in the form of ${{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( y \right)$ and on solving these equation with the help of trigonometric identities, we put the value of x and y as given in the question to solve this question. By putting the values in the identity and further solving it, we are able to get the option which follows our given equation.

Formula Used:
Identity which is used to solve this question is as follow:-
$\cos (x+y)=\cos x\cos y-\sin x\sin y$
And $\cos (x-y)=\cos x\cos y+\sin x\sin y$

Complete Step- by- step Solution:
Given that ${{\cos }^{2}}\left( \dfrac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \dfrac{\pi }{6}-\theta \right)$ …………………………… (1)
Let the given equation is in the form of ${{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( y \right)$
Where x = $\left( \dfrac{\pi }{6}+\theta \right)$ and y = $\left( \dfrac{\pi }{6}-\theta \right)$
We know the identity
$\cos (x+y)=\cos x\cos y-\sin x\sin y$
And $\cos (x-y)=\cos x\cos y+\sin x\sin y$
Then $\cos (x+y)\times \cos (x-y)=(\cos x\cos y-\sin x\sin y)\times (\cos x\cos y+\sin x\sin y)$
On further solving, we get
$\cos (x+y)\times \cos (x-y)=({{\cos }^{2}}x{{\cos }^{2}}y+\cos x\cos y\times \sin x\sin y-\sin x\sin y\times \cos x\cos y-{{\sin }^{2}}x{{\sin }^{2}}y)$
On cancelling the terms with opposite signs, we get $\cos (x+y)\times \cos (x-y)=({{\cos }^{2}}x{{\cos }^{2}}y-{{\sin }^{2}}x{{\sin }^{2}}y)$
Now we substitute ${{\cos }^{2}}y$ with $(1-{{\sin }^{2}}y)$ and ${{\sin }^{2}}x$ with $(1-{{\cos }^{2}}x)$
Then the equation becomes
$\cos (x+y)\times \cos (x-y)=({{\cos }^{2}}x(1-{{\sin }^{2}}y))-((1-{{\cos }^{2}}x){{\sin }^{2}}y)$
On expanding the above equation, we get
$\cos (x+y)\times \cos (x-y)=({{\cos }^{2}}x-{{\cos }^{2}}x{{\sin }^{2}}y-{{\sin }^{2}}y+{{\cos }^{2}}x{{\sin }^{2}}y)$
Simplify further, we get
$\cos (x+y)\times \cos (x-y)={{\cos }^{2}}x-{{\sin }^{2}}y$
Now on using the above identity in equation (1), we get
${{\cos }^{2}}\left( \dfrac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \dfrac{\pi }{6}-\theta \right)$ = $\cos \left( \dfrac{\pi }{6}+\theta +\dfrac{\pi }{6}-\theta \right)\cos \left( \dfrac{\pi }{6}+\theta -\dfrac{\pi }{6}+\theta \right)$………………….. (2)
Where x = $\left( \dfrac{\pi }{6}+\theta \right)$ and y = $\left( \dfrac{\pi }{6}-\theta \right)$
On further solving equation (2), we get
${{\cos }^{2}}\left( \dfrac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \dfrac{\pi }{6}-\theta \right)$= $\cos \left( \dfrac{2\pi }{6} \right)\cos \left( 2\theta \right)$
Then ${{\cos }^{2}}\left( \dfrac{\pi }{6}+\theta \right)-{{\sin }^{2}}\left( \dfrac{\pi }{6}-\theta \right)$= $\cos \left( \dfrac{\pi }{3} \right)\cos \left( 2\theta \right)$
We know value of $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$
Hence , ${{\cos }^{2}}\left( \dfrac{\pi }{4}-\beta \right)-{{\sin }^{2}}\left( \alpha -\dfrac{\pi }{4} \right)$ = $\dfrac{1}{2}\cos 2\theta $

Thus, Option ( A ) is correct.

Note: In this question, if we start to solve the question as with the given values of x and y, we get confused and it wastes our time. Try to solve these types of questions in the easiest way, so that we cannot get confused and solve the question in lesser time.