Question

Prove that :

\[ \sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = -1 \]

Solution

Reducing the angles,

\[ \sin\frac{10\pi}{3} = \sin\left(2\pi+\frac{4\pi}{3}\right) = \sin\frac{4\pi}{3} = -\frac{\sqrt3}{2} \]

\[ \cos\frac{13\pi}{6} = \cos\left(2\pi+\frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt3}{2} \]

\[ \cos\frac{8\pi}{3} = \cos\left(2\pi+\frac{2\pi}{3}\right) = \cos\frac{2\pi}{3} = -\frac12 \]

\[ \sin\frac{5\pi}{6} = \frac12 \]

Substituting these values,

\[ \begin{aligned} &\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} \\[8pt] =& \left(-\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right) + \left(-\frac12\right)\left(\frac12\right) \\[8pt] =& -\frac34-\frac14 \\[8pt] =& -1 \end{aligned} \]

Hence Proved.