Question

Find $\cos (\frac{\pi }{6}-A)\cos (\frac{\pi }{3}+B)-\sin (\frac{\pi }{6}-A)\sin (\frac{\pi }{3}+B)=$.

• A. $\sin (A-B)$
• B. $\sin (A+B)$
• C. $\cos (A-B)$
• D. $\cos (A+B)$

Answer 1

Answer: (A)

$\sin (A-B)$

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\cos (A+B)=\cos A\cos b-\sin A\sin B$

Using that here,

$\cos (\frac{\pi }{6}-A)\cos (\frac{\pi }{3}+B)-\sin (\frac{\pi }{6}-A)\sin (\frac{\pi }{3}+B)$

$\Rightarrow \cos (\frac{\pi }{6}-A+\frac{\pi }{3}+B)$

$\Rightarrow (\frac{\pi }{6}+\frac{\pi }{3}-(A-B\left)\right)$

$\Rightarrow \cos (\frac{\pi }{2}-(A-B\left)\right)$

$\Rightarrow \sin (A-B)$