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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds4 Marks
Question
Prove that: $\sin5\text{x}=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}$
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Answer

$\text{LHS}=\sin5\text{x}=\sin(3\text{x}+2\text{x})$ $=\sin3\text{x}\cos2\text{x}+\cos3\text{x}.\sin2\text{x}$ $=(3\sin\text{x}-4\sin^3\text{x})(1-2\sin^3\text{x})\\+(s\cos^3\text{x}=3\cos\text{x})2\sin\text{x}\cos\text{x}.$ $=3\sin\text{x}-4\sin^3\text{x}-6\sin^3\text{x}\\+8\sin^5\text{x}(8\cos^4\text{x}-6\cos^2\text{x})\sin\text{x}$ $=3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}\\+8\sin\text{x}\big((1-\sin^2\text{x})^2-6\sin\text{x}(1-\sin^2\text{x})\big)$ $=13\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+8\sin\text{x}\\-16\sin^3\text{x}8\sin^5\text{x}-6\sin\text{x}+6\sin^3\text{x}$ $=5\sin\text{x}-20\sin^3\text{x}+16\sin^5\text{x}=\text{RHS}$

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