Question

Find $a$ if $\sin (-{420}^{\circ })\left(\cos {390}^{\circ }\right)+\cos (-{660}^{\circ })\left(\sin {330}^{\circ }\right)=-a$

Answer 1

L.H.S.$=\sin (-{420}^{\circ })\left(\cos {390}^{\circ }\right)+\cos (-{660}^{\circ })\left(\sin {330}^{\circ }\right)$

$=-\sin {420}^{\circ }\cos {390}^{\circ }+\cos {660}^{\circ }\sin {330}^{\circ }$ $[\because \sin (-\theta )=-\sin \theta ,\cos (-\theta )=-\cos \theta ]$

$=-\sin ({90}^{\circ }\times 4+{60}^{\circ })\cos ({90}^{\circ }\times 4+{30}^{\circ })\cos ({90}^{\circ }\times 7+{30}^{\circ })\cos ({90}^{\circ }\times 3+{60}^{\circ })$

$=-\left(\sin {60}^{\circ }\right)\left(\cos {30}^{\circ }\right)+\left(\sin {30}^{\circ }\right)(-\cos {60}^{\circ })$

$=-\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}(-\frac{1}{2})=-1=$R.H.S.

Therefore, $a=1$
Ans: 1