Question

If \[\sin 2\theta + \sin 2\phi = \dfrac{1}{2}\] and \[\cos 2\theta + \cos 2\phi = \dfrac{3}{2}\] then \[{\cos ^2}\left( {\theta - \phi } \right) = \]
A) \[\dfrac{3}{8}\]
B) \[\dfrac{5}{8}\]
C) \[\dfrac{3}{4}\]
D) \[\dfrac{5}{4}\]

Answer 1

Hint: We are given trigonometric functions in sin and cos form and are asked to find the answer in \[{\cos ^2}x\] form. Here we will use the most common identity of trigonometry that is \[{\cos ^2}x + {\sin ^2}x = 1\] and the sum and difference formula to get the answer. Our first step will be squaring and adding both the sides of the given equations. Once we did it, we will proceed with the steps by using other trigonometric identities also.

Complete step by step solution:
Given that,
\[\sin 2\theta + \sin 2\phi = \dfrac{1}{2}\]…….equation (1)
\[\cos 2\theta + \cos 2\phi = \dfrac{3}{2}\]…..equation (2)
On squaring both the sides of equation (1) and equation (2),
\[{\left( {\sin 2\theta + \sin 2\phi } \right)^2} + {\left( {\cos 2\theta + \cos 2\phi } \right)^2} = {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{3}{2}} \right)^2}\]
On taking the whole square,
\[{\sin ^2}2\theta + 2\sin 2\theta \sin 2\phi + {\sin ^2}2\phi + {\cos ^2}2\theta + 2\cos 2\theta \cos 2\phi + {\cos ^2}2\phi = \dfrac{1}{4} + \dfrac{9}{4}\]
We know that, \[{\cos ^2}x + {\sin ^2}x = 1\]
Thus the above equation can be written as,
\[1 + 1 + 2\left( {\sin 2\theta \sin 2\phi + \cos 2\theta \cos 2\phi } \right) = \dfrac{5}{2}\]
\[2 + 2\left( {\sin 2\theta \sin 2\phi + \cos 2\theta \cos 2\phi } \right) = \dfrac{5}{2}\]
Taking 2 common,
\[2\left( {1 + \left( {\sin 2\theta \sin 2\phi + \cos 2\theta \cos 2\phi } \right)} \right) = \dfrac{5}{2}\]
On transposing 2 we get,
\[1 + \left( {\sin 2\theta \sin 2\phi + \cos 2\theta \cos 2\phi } \right) = \dfrac{5}{4}\]
\[1 + \cos \left( {2\theta - 2\phi } \right) = \dfrac{5}{4}\]
This can be written as,
\[1 + \cos 2\left( {\theta - \phi } \right) = \dfrac{5}{4}\]
As we know that \[1 + \cos 2\theta = 2{\cos ^2}\theta \] we can write the above equation as,
\[2{\cos ^2}\left( {\theta - \phi } \right) = \dfrac{5}{4}\]
Again transposing 2 we get,
\[{\cos ^2}x\left( {\theta - \phi } \right) = \dfrac{5}{8}\]
Thus this is the correct answer.
Thus option (B) is the correct answer.

Note:
Here note that, when we get the given equation we might lead with the sum and difference formulas but that is not the correct way because that will not simplify the equation rather. Also note that if the equation is written as \[{\sin ^2}\theta + {\sin ^2}\phi = \dfrac{1}{2}\& {\cos ^2}\theta + {\cos ^2}\phi = \dfrac{3}{2}\] then it will be very simple to solve as by just adding both the sides of the equation only.