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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds2 Marks
Question
Prove that: $(\sin^3\text{x}+\sin\text{x})\sin\text{x}+(\cos3\text{x}-\cos\text{x})\cos\text{x}=0$
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Answer

$\text{LHS}=(\sin3\text{A}+\sin\text{A})\sin\text{A}(\cos3\text{A}-\cos\text{A})\cos\text{A}$ $\Rightarrow2\sin2\text{A}.\cos\text{A}.\sin\text{A}+(-2\sin2\text{A}.\sin\text{A}\cos\text{A})$ $\begin{bmatrix}\because\sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}.\cos\frac{\text{C}-\text{D}}{2}\\\cos\text{C}-\cos\text{D}=-2\sin\frac{\text{C+D}}{2}.\sin\frac{\text{C+D}}{2}\end{bmatrix}$ $\Rightarrow2\sin2\text{A}.\cos\text{A}.\sin\text{A}-2\sin2\text{A}\cos\text{A}.\sin\text{A}$ $\Rightarrow0=\text{RHS}$

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