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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$\sin\text{x}+\sin3\text{x}+\sin5\text{x}+\sin7\text{x}=4\cos2\text{x}\sin4\text{x}\cos\text{x}$
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Answer

It is known that $\sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big).\cos\Big(\frac{\text{A}-\text{B}}{2}\Big).$$\text{L.H.S.}=\sin\text{x}+\sin3\text{x}+\sin5\text{x}+\sin7\text{x}$
$=(\sin\text{x}+\sin5\text{x})+(\sin3\text{x}+\sin7\text{x})$ $=2\sin\Big(\frac{\text{x}+5\text{x}}{2}\Big).\cos\Big(\frac{\text{x}-5\text{x}}{2}\Big)+2\sin\Big(\frac{3\text{x}+7\text{x}}{2}\Big)\cos\Big(\frac{3\text{x}-7\text{x}}{2}\Big)$ $=2\sin3\text{x}\cos(-2\text{x})+2\sin5\text{x}\cos(-2\text{x})$ $=2\sin3\text{x}\cos2\text{x}+2\sin5\text{x}\cos2\text{x}$ $=2\cos2\text{x}[\sin3\text{x}+\sin5\text{x}]$ $=2\cos2\text{x}\Big[2\sin\Big(\frac{3\text{x}+5\text{x}}{2}\Big).\cos\Big(\frac{3\text{x}-5\text{x}}{2}\Big)\Big]$ $=2\cos2\text{x}[2\sin4\text{x}.\cos(-\text{x})]$ $=4\cos2\text{x}\sin4\text{x}\cos\text{x}=\text{R.H.S.}$

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