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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
$\sin^3\text{x}+\sin^3\big(\frac{2\pi}{3}+\text{x}\big)+\sin^3\big(\frac{4\pi}{3}+\text{x}\big)=-\frac{3}{4}\sin3\text{x}$
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Answer

$\text{LHS}=\sin^3\text{x}+\sin^3\Big(\frac{2\pi}{3}+\text{x}\Big)+\sin^3\Big(\frac{4\pi}{3}+\text{x}\Big)$
$\Big\{$ We know that $\sin^3\text{x}=\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big\}$
$=\Big(\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big)+\Bigg\{\frac{3\sin\big(\frac{2\pi}{4}+\text{x}\big)-\sin3\big(\frac{2\pi}{3}+\text{x}\big)}{4}\Bigg\}\\+\Bigg\{\frac{3\sin\big(\frac{4\pi}{3}+\text{x}\big)-\sin3\big(\frac{4\pi}{3}+\text{x}\big)}{4}\Bigg\}$
$=\Big[\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big]+\Bigg[\frac{3\sin\big[\pi\big(\frac{2\pi}{3}+\text{x}\big)\big]-\sin(2\pi+3\text{x})}{4}\Bigg]\\+\Bigg[\frac{3\sin\big[\pi\big(\frac{\pi}{3}+\text{x}\big)\big]-\sin(4\pi+3\text{x})}{4}\Bigg]$
$=\frac{1}{4}\Big\{[3\sin\text{x}-\sin3\text{x}]+\Big[3\sin\Big(\frac{\pi}{3}-\text{x}\Big)-\sin3\text{x}\Big]\\-\Big[3\sin\Big(\frac{\pi}{3}+\text{x}\Big)+\sin3\text{x}\Big]\Big\}$
$=\frac{1}{4}\Big[3\sin\text{x}-\sin3\text{x}+3\sin\Big(\frac{\pi}{3}-\text{x}\Big)-3\sin\Big(\frac{\pi}{3}+\text{x}\Big)-\sin3\text{x}\sin3\text{x}\Big]$
$=\frac{1}{4}\Big[3\sin\text{x}-3\sin3\text{x}+3\Big(\sin\Big(\frac{\pi}{3}-\text{x}\Big)-\sin\Big(\frac{\pi}{3}+\text{x}\Big)\Big)\Big]$
$=\frac{1}{4}\Bigg[3\sin\text{x}-3\sin3\text{x}+3\Bigg\{2\cos\frac{\frac{\pi}{3}-\text{x}+\frac{\pi}{3}+\text{x}}{2}\sin\frac{\frac{\pi}{3}-\text{x}\frac{\pi}{3}-\text{x}}{2}\Bigg\}\Bigg]$
$=\frac{1}{4}\Big[3\sin\text{x}-3\sin2\text{x}+6\cos\frac{\pi}{3}\sin(-\text{x})\Big]$
$\frac{1}{4}[3\sin\text{x}-3\sin3\text{x}-3\sin\text{x}]$
$=-\frac{3}{4}\sin3\text{x}$
$=\text{RHS}$
$\text{LHS}=\text{RHS}$

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