Question

Prove that :

\[ \cos570^\circ\sin510^\circ+\sin(-330^\circ)\cos(-390^\circ)=0 \]

Solution

First reduce the angles.

\[ \cos570^\circ = \cos(360^\circ+210^\circ) = \cos210^\circ = -\frac{\sqrt3}{2} \]

\[ \sin510^\circ = \sin(360^\circ+150^\circ) = \sin150^\circ = \frac12 \]

Also,

\[ \sin(-330^\circ) = -\sin330^\circ = -\left(-\frac12\right) = \frac12 \]

\[ \cos(-390^\circ) = \cos390^\circ = \cos(360^\circ+30^\circ) = \cos30^\circ = \frac{\sqrt3}{2} \]

Substituting these values,

\[ \begin{aligned} &\cos570^\circ\sin510^\circ+\sin(-330^\circ)\cos(-390^\circ) \\[4pt] =& \left(-\frac{\sqrt3}{2}\times\frac12\right) + \left(\frac12\times\frac{\sqrt3}{2}\right) \\[4pt] =& -\frac{\sqrt3}{4}+\frac{\sqrt3}{4} \\[4pt] =& 0 \end{aligned} \]

Hence Proved.