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PART - 1 CH - 3 Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPART - 1 CH - 3 Trigonometric Functions3 Marks
Question
Prove that:
$4 \sin \alpha \cdot \sin \left(\alpha+\frac{\pi}{3}\right) \sin \left(\alpha+\frac{2 \pi}{3}\right)=\sin 3 \alpha$
✓

Answer


$\begin{array}{l} \text { L.H.S. } 4 \sin \alpha \sin \left(\alpha+\frac{\pi}{3}\right) \sin \left(\alpha+\frac{2 \pi}{3}\right) \\ = 2 \sin \alpha\left[2 \sin \left(\alpha+\frac{\pi}{3}\right) \cdot \sin \left(\alpha+\frac{2 \pi}{3}\right)\right] \\ = 2 \sin \alpha\left[\cos \left(\alpha+\frac{\pi}{3}-\alpha-\frac{2 \pi}{3}\right)-\cos \left(\alpha+\frac{\pi}{3}+\alpha+\frac{2 \pi}{3}\right)\right] \\ {[\because 2 \sin A \sin B =\cos ( A - B )-\cos ( A + B )] } \\ = 2 \sin \alpha\left[\cos \left(-\frac{\pi}{3}\right)-\cos (\pi+2 \alpha)\right] \\ = 2 \sin \alpha\left[\cos \frac{\pi}{3}+\cos 2 \alpha\right] \quad \left[\begin{array}{c}\because \cos (\pi+\theta)=-\cos \theta \\ \text { and } \cos (-\theta)=\cos \theta\end{array}\right]\ \\ = 2 \sin \alpha\left[\frac{1}{2}+\cos 2 \alpha\right] \end{array}$
$\begin{array}{l}=\sin \alpha+2 \sin \alpha \cos 2 \alpha \\ =\sin \alpha+[\sin (\alpha+2 \alpha)-\sin (2 \alpha-\alpha)]\end{array}$
$[\because 2 \cos A \cos B=\sin (A+B)-\sin (A-B)]$
$=\sin \alpha+\sin 3 \alpha-\sin \alpha=\sin 3 \alpha=$ R.H.S.

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