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3.Trigonometrical Ratios, Functions and Identities
medium
Prove that: $\sin x+\sin 3 x+\sin 5 x+\sin 7 x=4 \cos x \cos 2 x \sin 4 x$
Option A
Option B
Option C
Option D
Solution
It is known that $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)$
$L.H.S.$ $=\sin x+\sin 3 x+\sin 5 x+\sin 7 x$
$=(\sin x+\sin 5 x)+(\sin 3 x+\sin 7 x)$
$=2 \sin \left(\frac{x+5 x}{2}\right) \cdot \cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)$
$=2 \sin 3 x \cos (-2 x)+2 \sin 5 x \cos (-2 x)$
$=2 \sin 3 x \cos 2 x+2 \sin 5 x \cos 2 x$
$=2 \cos 2 x[\sin 3 x+\sin 5 x]$
$=2 \cos 2 x\left[2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)\right]$
$=2 \cos 2 x[2 \sin 4 x \cdot \cos (-x)]$
$=4 \cos 2 x \sin 4 x \cos x=R . H . S.$
Standard 11
Mathematics
