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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds3 Marks
Question
Prove that: $\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}=\frac{3}{4}\sin4\text{x}$
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Answer

$\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}=\frac{3}{4}\sin4\text{x}$ $\text{LHS}=\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}$ $=\Big(\frac{\cos3\text{x}+3\cos\text{x}}{4}\Big)\sin3\text{x}+\Big(\frac{3\sin\text{x}-\sin3\text{x}}{4}\Big)\cos3\text{x}$ $\begin{Bmatrix}\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\\\cos3\text{x}=4\cos^3\text{x}-3\cos\text{x}\end{Bmatrix}$ $=\frac{1}{4}[3(\sin3\text{x}\cos\text{x}+\sin\text{x}\cos3\text{x})\\+\cos3\text{x}\sin3\text{x}-\sin3\text{x}\cos3\text{x}]$ $=\frac{1}{4}[3\sin(3\text{x}+​​​​\text{x})+0]$ $\frac{3}{4}\sin4\text{x}$ So, $\cos^3\text{x}\sin3\text{x}+\sin^3\text{x}\cos3\text{x}=\frac{3}{4}\sin4\text{x}$ $=\text{RHS}$

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