How do you prove $\frac{sin x - 2 sin 2 x + sin 3 x}{sin x + 2 sin 2 x + sin 3 x} = - {tan}^{2} (\frac{x}{2})$ $(sinx-2sin2x+sin3x)/(sinx+2 sin2x+sin3x)= -tan^2(x/2)$ ?

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Answer 1 · 2018-06-08T00:13:25

$L H S = \frac{sin x - 2 sin 2 x + sin 3 x}{sin x + 2 sin 2 x + sin 3 x}$ $LHS=(sinx-2sin2x+sin3x)/(sinx+2 sin2x+sin3x)$

$= \frac{2 sin 2 x cos x - 2 sin 2 x}{2 sin 2 x cos x + 2 sin 2 x}$ $=(2sin2xcosx-2sin2x)/(2sin2xcosx+2 sin2x)$

$= - \frac{2 sin 2 x (1 - cos x)}{2 sin 2 x (1 + cos x)}$ $=-(2sin2x(1-cosx))/(2sin2x(1+cosx))$

$= - \frac{2 {sin}^{2} (\frac{x}{2})}{2 {cos}^{2} (\frac{x}{2})}$ $=-(2sin^2(x/2))/(2cos^2(x/2))$

$= - {tan}^{2} (\frac{x}{2}) = R H S$ $= -tan^2(x/2)=RHS$

How do you prove (sinx-2sin2x+sin3x)/(sinx+2 sin2x+sin3x)= -tan^2(x/2)sinx−2sin2x+sin3xsinx+2sin2x+sin3x=−tan2(x2)(sinx-2sin2x+sin3x)/(sinx+2 sin2x+sin3x)= -tan^2(x/2)?