Question

If $x=\cos \alpha +\cos \beta -\cos (\alpha +\beta )$ and $y=4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \left(\frac{\alpha +\beta }{2}\right)$, then $(x-y)$ equals.

• A. $0$
• B. $1$
• C. $-1$
• D. $-2$

Answer 1

The correct option is B $1$

$x=\cos \alpha +\cos \beta -\cos (\alpha +\beta )=2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}-(2{\cos }^2\frac{\alpha +\beta }{2}-1)$

$=1+2\cos \frac{\alpha +\beta }{2}(\cos \frac{\alpha -\beta }{2}-\cos \frac{\alpha +\beta }{2})$

$=1+4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\cos (\frac{\alpha +\beta }{2})$

$=1+y$

$⟹x-y=1$