Question

If $\sin \left(2a\right)+\sin \left(2b\right)=\frac{1}{2}$ and $\cos \left(2a\right)+\cos \left(2b\right)=\frac{3}{2}$, then ${\cos }^2\left(a-b\right)$ is equal to?

• A. $\frac{3}{8}$
• B. $\frac{5}{8}$
• C. $\frac{3}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is B

$\frac{5}{8}$

Explanation for the correct options:

Step $1$: Frame the equation.

Given:

$$\begin{gathered} \sin \left(2a\right)+\sin \left(2b\right)=\frac{1}{2} \\ \cos \left(2a\right)+\cos \left(2b\right)=\frac{3}{2} \end{gathered}$$

Since,$\sin \left(C\right)+\sin \left(D\right)=2\left[\sin \left(\frac{C+D}{2}\right)\right]\left[\cos \left(\frac{C-D}{2}\right)\right]$

$$\begin{gathered} \Rightarrow \sin \left(2a\right)+\sin \left(2b\right)=2\sin \left(a+b\right)\cos \left(a-b\right) \\ \Rightarrow 2\sin \left(a+b\right)\cos \left(a-b\right)=\frac{1}{2}.............\left(1\right) \end{gathered}$$

And

$$\begin{gathered} \Rightarrow \cos \left(C\right)+\cos \left(D\right)=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right) \\ \Rightarrow \cos \left(2a\right)+\cos \left(2b\right)=2\cos \left(a+b\right)\cos \left(a-b\right) \\ \Rightarrow 2\cos \left(a+b\right)\cos \left(a-b\right)=\frac{3}{2}.................\left(2\right) \end{gathered}$$

Step $2$: Compute the required value.

squaring and adding $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} \Rightarrow 4{\sin }^2\left(a+b\right){\cos }^2\left(a-b\right)+4{\cos }^2\left(a+b\right){\cos }^2\left(a-b\right)=\frac{9}{4}+\frac{1}{4} \\ \Rightarrow 4{\cos }^2\left(a-b\right)\left[{\sin }^2\left(a+b\right)+{\cos }^2\left(a+b\right)\right]=\frac{10}{4} \\ \Rightarrow {\cos }^2\left(a-b\right)=\frac{10}{16} \\ \Rightarrow {\cos }^2\left(a-b\right)=\frac{5}{8} \end{gathered}$$

Hence, option $\left(B\right)$ is the correct answer.