Question #7218e

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Answer 1 · 2016-04-16T21:29:43

See below

LHS=left hand side, RHS =right hand side

LHS= $\frac{sin (2 x + x)}{1 + 2 cos 2 x}$ $(sin(2x+x))/(1+2cos2x)$

$= \frac{sin 2 x cos x + cos 2 x sin x}{1 + 2 cos 2 x}$ $=(sin2xcosx+cos2xsinx)/(1+2cos2x)$

$= \frac{(2 sin x cos x) cos x + (1 - 2 {sin}^{2} x) sin x}{1 + 2 cos 2 x}$ $=((2sinxcosx)cosx+(1-2sin^2x)sinx)/(1+2cos2x)$

$= \frac{2 sin x {cos}^{2} x + sin x - 2 {sin}^{3} x}{1 + 2 (1 - 2 {sin}^{2} x)}$ $=(2sinxcos^2x+sinx-2sin^3x)/(1+2(1-2sin^2x))$

$= \frac{2 sin x (1 - {sin}^{2} x) + sin x - 2 {sin}^{3} x}{1 + 2 - 4 {sin}^{2} x}$ $=(2sinx(1-sin^2x)+sinx-2sin^3x)/(1+2-4sin^2x)$

$= \frac{2 sin x - 2 {sin}^{3} x + sin x - 2 {sin}^{3} x}{3 - 4 {sin}^{2} x}$ $=(2sinx-2sin^3x+sinx-2sin^3x)/(3-4sin^2x)$

$= \frac{3 sin x - 4 {sin}^{3} x}{3 - 4 {sin}^{2} x}$ $=(3sinx-4sin^3x)/(3-4sin^2x)$

$= \frac{sin x (3 - 4 {sin}^{2} x)}{3 - 4 {sin}^{2} x}$ $=(sinx(3-4sin^2x))/(3-4sin^2x)$

$= sin x$ $=sinx$

$= R H S$ $=RHS$