Question #4931b

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Question #4931b

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Answer 1 · 2017-04-17T21:16:44

$x = \frac{2}{9} y \sqrt{36 - y^{2}}$ $x= 2/9 ysqrt(36-y^2)$

Explanation:

The secret lies in using trig identities: $sin (2 t) = 2 sin (t) cos t (t); sin (t) = \sqrt{1 - {cos}^{2} (t)} .$ $sin(2t) = 2sin(t)cost(t); sin(t) = sqrt(1-cos^2(t)).$
We can write $cos (t) = \frac{y}{6}; and x = 4 \cdot 2 sin (t) cos (t) .$ $cos(t) = y/6; and x = 4*2sin(t)cos(t) .$
That is, $x = 8 cos (t) \sqrt{1 - {cos}^{2} (t)};$ $x = 8cos(t)sqrt(1-cos^2(t));$ than replace $cos (t)$ $cos(t)$ , so $x = 8 \cdot \frac{y}{6} \sqrt{1 - {(\frac{y}{6})}^{2}} = \frac{8}{36} y \sqrt{36 - y^{2}} = \frac{2}{9} y \sqrt{36 - y^{2}} .$ $x = 8*y/6sqrt(1-(y/6)^2) =8/36ysqrt(36-y^2) = 2/9ysqrt(36-y^2).$

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Question #4931b

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