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OpenStudy (anonymous):
quick question
is 1 - tan^2 x/1 + tan^2x = cos2x? Why is it so if yes?
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OpenStudy (anonymous):
@rational
OpenStudy (rational):
write tan^2 as sin^2/cos^2 and keep simplifying
OpenStudy (anonymous):
oh okay sorry.. another one is
cos^2x - sin^2x = 1 - tan x?
OpenStudy (rational):
\[\large \begin{align}
\dfrac{1-\tan^2x}{1+\tan^2x}&=\dfrac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x}}\\~\\
&= \dfrac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}\\~\\
&=\cdots
\end{align}\]
OpenStudy (anonymous):
I did not write down cos^2 x - sin^x/cos^x + sin^x formula :(
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OpenStudy (anonymous):
Is there any identity for root1+sinx + root1-sinx?
OpenStudy (rational):
\[\large \sqrt{1+\sin x} + \sqrt{1-\sin x}\]
like that?
OpenStudy (anonymous):
yeah!
OpenStudy (anonymous):
hang on lemme use the equation thing on openstudy.. sec
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OpenStudy (rational):
kk
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