Prove that: $sin 2 x + 2 sin 4 x + sin 6 x = 4 {cos}^{2} x sin 4 x$

Open in App

Solution

$sin 2 x + 2 sin 4 x + sin 6 x$

$= (sin 6 x + sin 2 x) + 2 sin 4 x$

$= 2 sin (\frac{6 x + 2 x}{2}) cos (\frac{6 x - 2 x}{2}) + 2 sin 4 x$

$= 2 sin 4 x cos 2 x + 2 sin 4 x$

$= 2 sin 4 x (cos 2 x + 1)$ ....... [ $∵ {cos}^{2} x = \frac{1 + cos 2 x}{2}$ ]

$= 2 sin 4 x 2 {cos}^{2} x$

$= 4 sin 4 x {cos}^{2} x$

Suggest Corrections

Similar questions

s i n 2 x + 2 s i n 4 x + s i n 6 x = 4 c o s^{2} x s i n 4 x

sin 2x + 2sin 4x + sin 6x = 4cos² x sin 4x

sin x 2 x + 2 sin 4 x + s i n 6 x = 4 {cos}^{2} x . sin 4 x

{sin}^{6} x + {cos}^{6} x = 1 - 3 {sin}^{2} x {cos}^{2} x

sin 2 x - sin 4 x + sin 6 x = 0

Join BYJU'S Learning Program