Mathematics
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OpenStudy (anonymous):
PROV THE IDENTITY
TAN^2Θ= ((1-cos(2Θ))/((1+cos(2Θ))
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OpenStudy (anonymous):
|dw:1393199938758:dw|
OpenStudy (anonymous):
@Mertsj can you tell me how to solve this
OpenStudy (anonymous):
anyone?
OpenStudy (helder_edwin):
try using
\[\large cos(2x)=\cos^2x-\sin^2x \]
it will follow easyly from this
OpenStudy (mertsj):
Going to use x instead of theta
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OpenStudy (anonymous):
ok
OpenStudy (mertsj):
\[\frac{1-\cos 2x}{1+\cos 2x}=\frac{1-(\cos ^2x-\sin ^2x)}{1+(\cos^2-\sin ^2x)}=\frac{1-\cos ^2x+\sin ^2x}{1-\sin ^2x+\cos ^2x}\]
OpenStudy (anonymous):
1st of all i used the wrong identity.. ok i get that part now though
OpenStudy (mertsj):
\[\frac{\sin ^2x+\sin ^2x}{\cos ^2x+\cos ^2x}=\frac{2\sin ^2x}{2\cos ^2x}\]
OpenStudy (anonymous):
okay. i understand that too
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OpenStudy (anonymous):
sin^2 is 1-cos^2 right? so would you do that next?
OpenStudy (mertsj):
What is 2/2 ?
OpenStudy (anonymous):
1. ok so it cancels
OpenStudy (mertsj):
What is:
\[\frac{\sin x}{\cos x}\]
OpenStudy (anonymous):
ohhh i understand. so youre left w/ sin^2x/ cos^2x which is tan^2x
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OpenStudy (mertsj):
yes
OpenStudy (anonymous):
thank you
OpenStudy (anonymous):
can you help me w/ one more
OpenStudy (anonymous):
i can post it as new question
OpenStudy (mertsj):
possibly
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OpenStudy (mertsj):
What is it?
OpenStudy (anonymous):
its another proving trig identity