Question

What is the value of $\sin {420}^o.\cos {390}^o+\cos (-{300}^o).\sin (-{330}^o)$ ?

• A. $0$
• B. $1$
• C. $-2$
• D. $-1$

Answer 1

The correct option is B $1$

Now,

$\cos (-\theta )=\cos \theta$

$\sin (-\theta )=-\sin \theta$

$\sin ({360}^o+\theta )=\sin \theta$

$\cos ({360}^o+\theta )=\cos \theta$

$\sin ({360}^o-\theta )=-\sin \theta$

$\cos ({360}^o-\theta )=\cos \theta$

$\therefore \sin {420}^o\cos {390}^o+\cos (-{300}^o).\sin (-{330}^o)$

$=\sin ({360}^o+{60}^o).\cos ({360}^o+{30}^o)-\cos ({360}^o-{60}^o).\sin ({360}^o-{30}^o)$

$=\sin {60}^o.\cos {30}^o+\cos {60}^o\sin {30}^o$

$=\frac{3}{4}+\frac{1}{4}=1$

Hence, B is correct.