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

જો $0 < x , y < \pi$ અને $\cos x +\cos y-\cos ( x + y )=\frac{3}{2}$ હોય, તો $\sin x+\cos y =$ ...... .

Similar Questions

$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}=...............$

$2\cos x - \cos 3x - \cos 5x = $

$\frac{{\sec \,8\theta - 1}}{{\sec \,4\theta - 1}}$ =

${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $