Question

The value of $ \sin {{600}^{o}}\,\cos {{330}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}} $ is:

• A. 1
• B. $ -1 $
• C. $ \frac{\sqrt{3}}{2} $
• D. $ \frac{1}{\sqrt{2}} $

Answer 1

Answer: (B)

$ \sin {{600}^{o}}\cos {{330}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}} $
$ =\sin ({{720}^{o}}-{{120}^{o}})\cos ({{180}^{o}}+{{150}^{o}}) $
$ +\cos {{120}^{o}}\sin {{150}^{o}} $ $ =\sin {{120}^{o}}\cos {{150}^{o}}+\cos {{120}^{o}}\sin {{150}^{o}} $
$ =\sin ({{120}^{o}}+{{150}^{o}}) $
$ =\sin {{270}^{o}}=\sin ({{180}^{o}}+{{90}^{o}}) $
$ =-\sin {{90}^{o}}=-1 $