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) =
(a) $\frac{3}{8}$

(b) $\frac{5}{8}$

(c) $\frac{3}{4}$

(d) $\frac{5}{4}$

Answer 1

(b) $\frac{5}{8}$
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 = $\frac{1}{4}+\frac{9}{4}$

$$\begin{gathered} \Rightarrow {\left[2\sin \left(\frac{2\theta +2\varphi }{2}\right)\cos \left(\frac{2\theta -2\varphi }{2}\right)\right]}^2+{\left[2\cos \left(\frac{2\theta +2\varphi }{2}\right)\cos \left(\frac{2\theta -2\varphi }{2}\right)\right]}^2=\frac{5}{2} \\ \Rightarrow 4{\sin }^2\left(\theta +\varphi \right){\cos }^2\left(\theta -\varphi \right)+4{\cos }^2\left(\theta +\varphi \right){\cos }^2\left(\theta -\varphi \right)=\frac{5}{2} \\ \Rightarrow 4{\cos }^2\left(\theta -\varphi \right)\left[{\sin }^2\left(\theta +\varphi \right)+{\cos }^2\left(\theta +\varphi \right)\right]=\frac{5}{2} \\ \Rightarrow 4{\cos }^2\left(\theta -\varphi \right)=\frac{5}{2} \\ \Rightarrow {\cos }^2\left(\theta -\varphi \right)=\frac{5}{8} \end{gathered}$$