Question #0d91d

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Question #0d91d

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Answer 1 · 2016-06-17T14:40:51

$y = (\frac{3}{4}) (2 - x^{2}) .$ $y= (3/4)(2-x^2).$

Explanation:

Recall the identity : ${sin}^{2} θ = \frac{1 - cos 2 θ}{2} .$ $sin^2theta=(1-cos2theta)/2.$

Hence,
$y = 3 {sin}^{2} θ = (\frac{3}{2}) (1 - cos 2 θ) .$ $y=3sin^2theta=(3/2)(1-cos2theta).$ ..............(1)

But, it is given that $x = \sqrt{2 cos 2 θ},$ $x=sqrt(2cos2theta),$
so that $\frac{x^{2}}{2} = cos 2 θ .$ $x^2/2=cos2theta.$

Now, putting this value of $cos 2 θ$ $cos2theta$ in (1), we get,

$y = (\frac{3}{2}) (1 - \frac{x^{2}}{2}) = (\frac{3}{4}) (2 - x^{2}) .$ $y= (3/2)(1-x^2/2)=(3/4)(2-x^2).$

Answer 2 · 2016-06-17T15:08:39

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

Explanation:

$y = 3 {sin}^{2} θ$ $y=3sin^2theta$
$x = \sqrt{2 cos 2 θ}$ $x=sqrt(2cos2theta)$
$x^{2} = 2 cos 2 θ$ $x^2=2cos2theta$
= $2 {cos}^{2} θ - 2 {sin}^{2} θ$ $2cos^2theta-2sin^2theta$
= $2 {cos}^{2} θ - \frac{2}{3} y$ $2cos^2theta-2/3y$
= $2 (1 - \frac{1}{3} y) - \frac{2}{3} y$ $2(1-1/3y)-2/3y$
= $2 - \frac{4}{3} y$ $2-4/3y$
so
$y = - \frac{3}{4} (x^{2} - 2)$ $y=-3/4(x^2-2)$

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Question #0d91d

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