Let $\alpha, \beta$ be two real number such that $\pi

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Let $\alpha, \beta$ be two real number such that $\pi < (\alpha-\beta) < 3 \pi$. If $\sin \alpha+\sin \beta=\frac{-21}{65}$ and $\cos \alpha+\cos \beta=\frac{-2}{65}$, then $\cos \left(\frac{\beta-\alpha}{2}\right)=$

Aicharya · Accepted Answer

$(\sin \alpha+\sin \beta)^2=\left(\frac{-21}{65}ight)^2$ $\sin ^2 \alpha+\sin ^2 \beta+2 \sin \alpha \sin \beta=\frac{441}{(65)^2}$ ...(i) $(\cos \alpha+\cos \beta)^2=\left(\frac{-2}{65}ight)^2$ $\cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta=\frac{4}{(65)^2}$ ...(ii) Eqn. (i) + Eqn. (ii) $\begin{aligned} & 2+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)=\frac{445}{4225}=\frac{89}{845} \ & 2 \cos (\alpha-\beta)=\frac{89}{845}-2=-\frac{1601}{845} \ & \cos (\alpha-\beta)=-\frac{1601}{1690}\end{aligned}$ $\begin{aligned} & 2 \cos ^2\left(\frac{\alpha-\beta}{2}ight)-1=\frac{-1601}{1690} \ & 2 \cos ^2\left(\frac{\alpha-\beta}{2}ight)=\frac{89}{1690} \ & \cos ^2\left(\frac{\alpha-\beta}{2}ight)=\frac{89}{1690 	imes 2} \ & \cos \left(\frac{\alpha-\beta}{2}ight)=-\sqrt{\frac{89}{1690 	imes 2}}\end{aligned}$ $\begin{aligned} & \pi < \alpha-\beta < 3 \pi \ & \frac{\pi}{2} < \frac{\alpha-\beta}{2} < \frac{3 \pi}{2}\end{aligned}$ $\cos (\alpha-\beta)$ is negative $\begin{aligned} & 	herefore \cos \left(\frac{\alpha-\beta}{2}ight)=-\frac{\sqrt{89}}{26 \sqrt{5}} \ & \cos \left(\frac{\beta-\alpha}{2}ight)=\frac{-\sqrt{89}}{26 \sqrt{5}}(\because \cos 	heta=\cos (-	heta))\end{aligned}$ Note: The correct option according to key is possible if $\cos \alpha+\cos \beta=-27 / 65$.

Let $\alpha, \beta$ be two real number such that $\pi < (\alpha-\beta) < 3 \pi$. If $\sin \alpha+\sin…

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