Question
If $(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\lambda\cos^2\Big(\frac{\alpha-\beta}{2}\Big),$ write the value of $\lambda.$

Answer

We have, $\text{LHS}=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$ $=\ \Big[2\cos\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)\Big]^2+\Big[2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)\Big]^2$ $=\ 4\cos^2\Big(\frac{\alpha+\beta}{2}\Big)\cos^2\Big(\frac{\alpha-\beta}{2}\Big)+4\sin^2\Big(\frac{\alpha+\beta}{2}\Big)\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$ $=\ 4\cos^2\Big(\frac{\alpha-\beta}{2}\Big)\Big[\cos^2\Big(\frac{\alpha+\beta}{2}\Big)+\sin^2\Big(\frac{\alpha+\beta}{2}\Big)\Big]$ $=\ 4\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$ $[\because\ \cos^2\theta+\sin^2\theta=1]$ $\Rightarrow\ (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=4\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$ It is given that, $(\cos\alpha+\cos\beta)^2+(\sin\alpha+\cos\beta)^2=\lambda\cos^2\Big(\frac{\alpha-\beta}{2}\Big)$ On comparing, we get $\lambda=4$ $\therefore\ \lambda=4$

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