यदि 0

Question

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$

$\cos ^{2}\left(\frac{x+y}{2}\right)-\cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)$

Vedclass · Accepted Answer

$\frac{1+\sqrt{3}}{2}$

Vedclass · Answer

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$

$\cos ^{2}\left(\frac{x+y}{2}ight)-\cos \left(\frac{x+y}{2}ight) \cdot \cos \left(\frac{x-y}{2}ight)$

$+\frac{1}{4} \cdot \cos ^{2}\left(\frac{x-y}{2}ight)+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}ight)=0$

$\Rightarrow\left(\cos \left(\frac{x+y}{2}ight)-\frac{1}{2} \cos \left(\frac{x-y}{2}ight)ight)^{2}+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}ight)=0$

$\Rightarrow \sin \left(\frac{x-y}{2}ight)=0$ and $\cos \left(\frac{x+y}{2}ight)=\frac{1}{2} \cos \left(\frac{x-y}{2}ight)$

$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$

$	herefore \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

यदि $0 < x, y < \pi$ तथा $\cos x+\cos y-\cos (x+y)=\frac{3}{2}$, है, तो $\sin x+\cos y$ बराबर है

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