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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions2 Marks
Question
Prove that:
$\sin\frac{13\pi}{3}\sin\frac{8\pi}{3}+\cos\frac{2\pi}{3}\sin\frac{5\pi}{6}=\frac{1}{2}$
✓

Answer

$\text{L.H.S}=\sin\frac{13\pi}{3}\sin\frac{8\pi}{3}+\cos\frac{2\pi}{3}\sin\frac{5\pi}{6}$
$=\sin780^\circ\sin480^\circ+\cos120^\circ\sin150^\circ$
$=\sin\Big(4\pi+\frac{\pi}{3}\Big)\sin\Big(3\pi+\frac{\pi}{3}\Big)+\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)\sin\Big(\pi-\frac{\pi}{6}\Big)$ $(\because\pi=180^\circ)$
$=\sin\frac{\pi}{3}\times\sin\frac{\pi}{3}+\Big(-\sin\frac{\pi}{6}\Big)\sin\frac{\pi}{6}$ $\begin{pmatrix}\because\sin\Big(4\pi+\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\\\&\sin\Big(3\pi-\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\end{pmatrix}$
$=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac{1}{2}\times\frac{1}{2}$
$=\frac{3}{4}-\frac{1}{4}$
$=\frac{2}{4}$
$=\frac{1}{2}$
$=\text{R.H.S}$
$\text{Proved}$

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