Question

show that
sin 150 cos 120 +cos 330 sin 660 = -1

Answer 1

We know,

$\sin 150^o=\sin (180^o-30^o)=\sin (30^o)=0.5\\ \cos 120^o=-\cos 60^o=-0.5\\ \cos 330^o=\cos (360^o-30^o)=\cos (30^o)=\frac{\sqrt{3}}{2} \\ \sin 660^o=\sin (720^o-60^o)=-\sin (60^o)=-\frac{\sqrt{3}}{2}$

Now plugging the above values in the original equation,

$\sin 150^o\cos 120^o+\cos 330^o\sin 660^o=-(\frac{1}{2})^2-(\frac{\sqrt{3}}{2} )^2=-1=RHS$

Answer 2

Answer and Explanation:

To show : $\sin 150\times \cos120+\cos 330\times \sin 660=-1$

Solution :  

Take LHS,

$\sin 150\times \cos120+\cos 330\times \sin 660$

Applying trigonometry we can write term as,

$\sin 660=\sin(720-60)=-\sin(60)$

$\cos 330=\cos(360-30)=\cos(30)$

$\cos 120=\cos(180-60)=-\cos(60)$

$\sin 150=\sin(180-30)=\sin(30)$

Substitute the values in the expression,

$=\sin(30)\times(-\cos(60))-\cos(60)\times (-\sin(60))$

Put all the values,

$=- \frac{1}{2}\times\frac{1}{2}-\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}$

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

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

$=\frac{-4}{4}$

$=-1$

=RHS

Therefore, $\sin 150\times \cos120+\cos 330\times \sin 660=-1$