Question #58780

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Answer 1 · 2017-11-01T00:48:07

$R H S = \frac{{sin}^{2} x}{2 + sin 2 x csc x}$ $RHS=(sin^2x)/(2+sin2xcscx)$

$= \frac{{(2 sin (\frac{x}{2}) cos (\frac{x}{2}))}^{2}}{2 + 2 sin x cos x \times \frac{1}{sin x}}$ $=(2sin(x/2)cos(x/2))^2/(2+2sinxcosx xx1/sinx)$

$= \frac{4 {sin}^{2} (\frac{x}{2}) {cos}^{2} (\frac{x}{2})}{2 (1 + cos x)}$ $=(4sin^2(x/2)cos^2(x/2))/(2(1+cosx ))$

$= \frac{4 {sin}^{2} (\frac{x}{2}) {cos}^{2} (\frac{x}{2})}{2 \cdot 2 {cos}^{2} (\frac{x}{2})}$ $=(4sin^2(x/2)cos^2(x/2))/(2*2cos^2(x /2))$

$= {sin}^{2} (\frac{x}{2}) = L H S$ $=sin^2(x/2)=LHS$

Answer 2 · 2017-11-01T17:18:01

Formulas to be used in the proof.

$sin 2 x = 2 sin x cos x$ $sin 2x = 2 sin x cos x$
$sin (\frac{1}{2} x) = \pm \sqrt{\frac{1 - cos x}{2}}$ $sin (1/2 x)=+-sqrt((1-cosx)/2)$
$:.sin ^2(1/2 x)=(1-cosx)/2$
$cscx=1/sinx$
$sin^2x+cos^2x=1$
$sin^2x=1-cos^2x$

Right Hand Side:

$sin^2x/(2+sin2xcscx)=sin^2x/(2+2sinx cosx cscx)$

$=sin^2x/(2+2cancelsinx cosx * 1/cancelsinx)$

$=(1-cos^2x)/(2+2cosx )$

$=((1-cosx)(1+cosx))/(2(1+cosx ))$

$=((1-cosx)cancel(1+cosx))/(2cancel(1+cosx ))$

$=(1-cosx)/2$

$=sin^2(1/2)x$

$:.=$ Left Hand Side