Prove that: 3A+ 5A+ 7A+ 15A=4 4A 5A 6A - Vidyadip

Question

Prove that:
$\cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos15\text{A}=4\cos4\text{A}\cos5\text{A}\cos6\text{A}$

✓

Answer

We have,
$\text{LHS}=\cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos15\text{A}$
$=\ [\cos5\text{A}+\cos3\text{A}]+[\cos15\text{A}+\cos7\text{A}]$
$=\ \Big[2\cos\frac{5\text{A}+3\text{A}}{2}\cos\frac{5\text{A}-3\text{A}}{2}\Big]+\Big[2\cos\frac{15\text{A}+7\text{A}}{2}\cos\frac{15\text{A}-7\text{A}}{2}\Big]$
$=\ 2\cos4\text{A}\cos\text{A}+2\cos11\text{A}\cos4\text{A}$
$=\ 2\cos4\text{A}[\cos\text{A}+\cos11\text{A}]$
$=\ 2\cos4\text{A}[\cos11\text{A}+\cos\text{A}]$
$=\ 2\cos4\text{A}\Big[2\cos\frac{(11\text{A}+\text{A)}}{2}\cos\frac{(11\text{A}-\text{A})}{2}\Big]$
$=\ 4\cos\text{A}[\cos6\text{A}\cos5\text{A}]$
$=\ 4\cos4\text{A}\cos5\text{A}\cos6\text{A}$
$=\ \text{RHS}$
$\therefore\ \cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos15\text{A}=4\cos4\text{A}\cos5\text{A}\cos6\text{A}$
Hence proved.

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