Question

If $0<\alpha ,\beta <\pi$ and $cos\alpha +cos\beta -cos(\alpha +\beta )=3/2$ then $984\sqrt{3}sin\alpha +896cos\alpha$ is equal to

Answer 1

Given $\cos \alpha +\cos \beta -\cos (\alpha +\beta )=\frac{3}{2}$
$\Rightarrow 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}-(2{\cos }^2\frac{\alpha +\beta }{2}-1)=\frac{3}{2}$
$\Rightarrow 4{\cos }^2\frac{\alpha +\beta }{2}-4\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+1=0$
$\Rightarrow {[2\cos \frac{\alpha +\beta }{2}-\cos \frac{\alpha -\beta }{2}]}^2+{\sin }^2\frac{\alpha -\beta }{2}=0$
$\Rightarrow \sin \frac{\alpha -\beta }{2}=0$ and $2\cos \frac{\alpha +\beta }{2}=\cos \frac{\alpha -\beta }{2}$
$\Rightarrow \alpha =\beta$ and $2\cos \alpha =1$
$\Rightarrow \alpha =\frac{\pi }{3}(0<\alpha ,\beta <\pi )$
$\Rightarrow \sin \alpha =\frac{\sqrt{3}}{2},\cos \alpha =\frac{1}{2}$
So, $984\sqrt{3}\sin \alpha +896\cos \alpha =1924$.