Question

If sin 2 θ + sin 2 ϕ = \[\frac{1}{2}\] and cos 2 θ + cos 2 ϕ = \[\frac{3}{2}\], then cos² (θ − ϕ) =

 

 

Options

• \[\frac{3}{8}\]

   

• \[\frac{5}{8}\]

   

• \[\frac{3}{4}\]

   

• \[\frac{5}{4}\]

   

Answer 1

\[\frac{5}{8}\]
Given:
sin 2θ + sin 2ϕ = \[\frac{1}{2}\]                  .....(i)
and
cos 2θ + cos 2ϕ = \[\frac{3}{2}\]          .....(ii)
Squaring and adding (i) and (ii), we get:
(sin 2θ + sin 2ϕ)² + (cos 2θ + cos 2ϕ)² = \[\frac{1}{4} + \frac{9}{4}\]
\[\Rightarrow \left[ 2\sin\left( \frac{2\theta + 2\phi}{2} \right)\cos\left( \frac{2\theta - 2\phi}{2} \right) \right]^2 + \left[ 2\cos\left( \frac{2\theta + 2\phi}{2} \right)\cos\left( \frac{2\theta - 2\phi}{2} \right) \right]^2 = \frac{5}{2}\]
\[ \Rightarrow 4 \sin^2 \left( \theta + \phi \right) \cos^2 \left( \theta - \phi \right) + 4 \cos^2 \left( \theta + \phi \right) \cos^2 \left( \theta - \phi \right) = \frac{5}{2}\]
\[ \Rightarrow 4 \cos^2 \left( \theta - \phi \right)\left[ \sin^2 \left( \theta + \phi \right) + \cos^2 \left( \theta + \phi \right) \right] = \frac{5}{2}\]
\[ \Rightarrow 4 \cos^2 \left( \theta - \phi \right) = \frac{5}{2}\]
\[ \Rightarrow \cos^2 \left( \theta - \phi \right) = \frac{5}{8}\]