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Trigonometric Ratios for Sum of Two Angles

Prove that si...

Question

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

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Solution

$L . H . S . = sin 2 x + 2 sin 4 x + sin 6 x$
$= (sin 6 x + sin 2 x) + 2 sin 4 x$
$sin A + sin B = 2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})$
Let $A = 2 x, B = 6 x$
$\Rightarrow L . H . S = 2 sin (\frac{2 x + 6 x}{2}) cos (\frac{2 x - 6 x}{2}) + 2 sin 4 x$
$= 2 sin (4 x) cos (2 x) + 2 sin 4 x$
$= 2 sin (4 x) (1 + cos (2 x)) = 2 sin (4 x) (2 {cos}^{2} x)$
$= 4 {cos}^{2} x sin 4 x = R . H . S$
Hence Proved.

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Trigonometric Ratios for Sum of Two Angles

Standard XII Mathematics

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