If ( + )^2+( + )^2= ^2 ( - 2 ), write the value of . - Vidyadip

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 $\lambda.$

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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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