Question #c2ea6

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Question #c2ea6

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Answer 1 · 2018-02-14T05:26:52

Future reference:-(see explanation first)

From, $(2)$ $(2)$ and $(3)$ $(3)$ ,

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

Explanation:

$sin x + {sin}^{2} x + {sin}^{3} x = 1$ $sinx+sin^2x+sin^3x=1$

$=>sinx+sin^3x=1-sin^2x" "...(2)$

$=>sinx(1+sin^2x)=cos^2x" "...(3)$

$=>sin^2x(1+sin^2x)^2=(cos^2x)^2$

$=>(1-cos^2x)(2-cos^2x)^2=cos^4x$

$=>(1-cos^2x)(4-4cos^2x+cos^4x)=cos^4x$

$=>4-4cos^2x+cos^4x-4cos^2x+4cos^4x-cos^6x=cos^4x$

$=>4-4cos^2x+cancel(cos^4x)-4cos^2x+4cos^4x-cos^6x=cancel(cos^4x$

$=>4-4cos^2x-4cos^2x+4cos^4x-cos^6x=0$

$=>4(1-2cos^2x+cos^4x)=cos^6x$

$=>4(1-cos^2x)^2=cos^6x$

$=>2(1-cos^2x)=cos^3x$

$=>cos^3x+cos^2x=2-cos^2x$

$=>cos^3x+cos^2x+cosx=2-cos^2x+cosx$

$=>cos^3x+cos^2x+cosx=1+sin^2x+(2sinx)/(1+sin^2x)$

$=>cos^3x+cos^2x+cosx=((1+sin^2x)^2+2sinx)/(1+sin^2x)$

$=>cos^3x+cos^2x+cosx=$

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Question #c2ea6

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