Question #cc719

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Question #cc719

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Answer 1 · 2016-07-24T10:09:16

${sin}^{2} 2 x ({cot}^{2} x - {tan}^{2} x) = 4 cos 2 x$ $sin^2 2x(cot^2x-tan^2x)=4cos2x$
$= sin 2 x \cdot sin 2 x ({cot}^{2} x - {tan}^{2} x)$ $=sin2x*sin2x(cot^2x-tan^2x)$

Using the double angle formula for sin and converting cot and tan to terms with sin and cos we now have

$= 4 {sin}^{2} x {cos}^{2} x (\frac{{cos}^{2} x}{{sin}^{2} x} - \frac{{sin}^{2} x}{{cos}^{2} x})$ $=4sin^2xcos^2x(cos^2x/sin^2x - sin^2x/cos^2x)$
$= 4 {sin}^{2} x {cos}^{2} x (\frac{{cos}^{4} x - {sin}^{4} x}{{sin}^{2} x {cos}^{2} x})$ $=4sin^2xcos^2x((cos^4x-sin^4x) / (sin^2xcos^2x))$
$= 4 {sin}^{2} x {cos}^{2} x \frac{({cos}^{2} x + {sin}^{2} x) ({cos}^{2} x - {sin}^{2} x)}{({sin}^{2} x {cos}^{2} x)}$ $=4sin^2xcos^2x((cos^2x +sin^2x)(cos^2x-sin^2x))/((sin^2xcos^2x))$
$= 4 {sin}^{2} x {cos}^{2} x (\frac{{cos}^{2} x - {sin}^{2} x}{{sin}^{2} x \cdot {cos}^{2} x})$ $=4sin^2xcos^2x((cos^2x-sin^2x)/(sin^2x*cos^2x))$ since ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x +sin^2x =1$
$= \frac{4 {cos}^{4} {sin}^{2} x - {sin}^{4} x {cos}^{2} x}{{sin}^{2} x {cos}^{2} x}$ $=(4cos^4sin^2x - sin^4xcos^2x)/(sin^2xcos^2x)$
$=(4cancel(sin^2xcos^2x)(cos^2x-sin^2x))/(cancel(sin^2xcos^2x))$
$=4(cos^2x-sin^2x)$
$=4cos2x$ since $cos^2x-sin^2x=cos2x$

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