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July 30, 2014

July 30, 2014

Posted by **dreidel** on Wednesday, December 18, 2013 at 10:34pm.

x=t+1

y = 2t^2 +1

for t in the set I

is identical to the graph of the parametric equations

x = 1+ sin s

y = 2- cos(2s)

for s in [0,pi)?

- precalculus -
**Steve**, Thursday, December 19, 2013 at 12:25amit appears we need

1+t = 1+sin s,so

t = sin s

t must be in [0,1] if s is in [0,pi)

also,

2t^2+1 = 2-cos(2s)

2t^2+1 = 2 - (1-2sin^2 s)

2t^2+1 = 2sin^2 s + 1

same domain.

- precalculus -
**dreidel**, Thursday, December 19, 2013 at 6:05pmThx!

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