If 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}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

If $0 < x , y < \pi$ and $\cos x +\cos y-\cos ( x + y )=\frac{3}{2},$ then $\sin x+\cos y$ is equal to ...... .

Similar Questions

The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$

If $\tan x + \tan \left( {\frac{\pi }{3} + x} \right) + \tan \left( {\frac{{2\pi }}{3} + x} \right) = 3,$ then

$A, B, C$ are the angles of a triangle, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\,\cos B\,\cos C = $

If $x + y + z = {180^o},$ then $\cos 2x + \cos 2y - \cos 2z$ is equal to