Sbdenney
HELP! A simplified version of sin2 θ (1 + cot2 θ) = 1 is



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mwmnj
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hmm, dunno beside just distributing

mwmnj
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dint see a trig sub

Sbdenney
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:/

Sbdenney
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trig sub?

Tyler1992
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i get just \[\sin^2\theta + cox^2 \theta\]

mwmnj
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How?

Sbdenney
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Whats that Tyler?

Sbdenney
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whats the 2nd part of it

mwmnj
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Oh wait actually

mwmnj
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divide sin^2

Tyler1992
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I did \[\sin^2\theta + \sin^2\theta*\frac{\cos^2\theta}{\sin^2\theta}\]

Sbdenney
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sin^2 θ (csc^2 θ) ?
is that what you meant?

mwmnj
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then you have 1+cos^2=1/sin^2

mwmnj
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which is sin^2=1/sin^2

mwmnj
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so then you multiply sin^2 back to have sin^4 = 1

Sbdenney
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ALl my answers start with sin2 θ

Tyler1992
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Then you get sin^2 + cos^2 which is 1

DHASHNI
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sin^2 θ (1 + cot^2 θ) = 1
sin^2 θ + cos^2 θ = 1

Sbdenney
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:(

Tyler1992
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??

Tyler1992
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\[Answer = \sin^2\theta + \cos^2 \theta\]