Can someone help me finish this problem I took it as far as I could... If x=9sinθ, use trigonometric substitution to write √(81-x^2) as a trigonometric function of θ, where -pi/2<θ

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Can someone help me finish this problem I took it as far as I could... If x=9sinθ, use trigonometric substitution to write √(81-x^2) as a trigonometric function of θ, where -pi/2<θ

Mathematics
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So far I have √(81-9sinθ^2) 81√(1-sinθ^2) 1-sin^2θ=1/cscθ ?
or maybe It is 9√(1-sinθ^2)...
yes

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Other answers:

your second post was right
√(81-81sinθ^2) <--- (9sinθ)^2 = 81sinθ^2 = 9√(1-sinθ^2)
next you need to figure out which trig substitution to use
Thank you! would the answer be 9cscθ then?
\[\sin ^{2}\theta+\cos ^{2}\theta = 1\] is a where you want to start thinking about your substitution
the answer isn't 9cscθ, but you are almost there
use \[1-\sin ^{2}\theta = \cos ^{2}\theta\]
oh ok...where did the cos come from? I thought sinθ=1/cscθ so in this case it would be 1-sin^2θ=1/cscθ ?
ok so 9cosθ?
yes :)
Sweet! thanks a bunch!

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