Prove that: + 3A+ 5A+ 7A=4 2A 4A - Vidyadip

Question

Prove that: $\cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}=4\cos\text{A}\cos2\text{A}\cos4\text{A}$

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Answer

We have, $\text{LHS}=\cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}$ $=\ [\cos3\text{A}+\cos\text{A}]+[\cos7\text{A}+\cos5\text{A}]$ $=\ \Big[2\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}-\text{A}}{2}\Big)\Big]+\Big[2\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)\Big]$ $=\ 2\cos2\text{A}\cos\text{A}+2\cos6\text{A}\cos4\text{A}$ $=\ 2\cos\text{A}[\cos2\text{A}+\cos6\text{A}]$ $=\ 2\cos\text{A}[\cos2\text{A}+\cos6\text{A}]$ $=\ 2\cos2\text{A}\Big[2\cos\Big(\frac{6\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{6\text{A}-\text{A}}{2}\Big)\Big]$ $=\ 4\cos\text{A}[\cos4\text{A}\cos2\text{A}]$ $=\ \text{RHS}$ $\therefore\ \cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}=4\cos\text{A}\cos2\text{A}\cos4\text{A}$ Hence proved.

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