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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: $3\text{x}+\sin2\text{x}-\sin\text{x}=4\sin\text{x}\cos\frac{\text{x}}{2}\cos\frac{3\text{x}}{2}$
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Answer

$\sin3\text{x}+\sin2\text{x}-\sin\text{x}$ $=(\sin3\text{x}-\sin\text{x})+\sin2\text{x}$ $=2\cos\Big(\frac{3\text{x}+\text{x}}{2}\Big)\sin\Big(\frac{3\text{x}-\text{x}}{\text{2}}\Big)+2\sin\text{x}\cos\text{x}$ $\begin{bmatrix}\because\sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\\\sin2\text{x}=2\sin\text{x}\cos\text{x}\end{bmatrix}$ $=2\cos(2\text{x})\sin\text{x}+2\sin\text{x}\cos\text{x}$ $=2\sin\text{x}[\cos(2\text{x})+\cos\text{x}]$ $=2\sin\text{x}\Big[2\cos\Big(\frac{2\text{x}+\text{x}}{2}\Big)+\cos\Big(\frac{2\text{x}-\text{x}}{2}\Big)\Big]$ $\Big[\because\cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$ $=4\sin\text{x}\cos\frac{3\text{x}}{2}\cos\frac{\text{x}}{2}$

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