Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. angle
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pka
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First of all it has to be \(\displaystyle \left[ {\tan (x)} \right]^2 = \tan ^2 (x).\)
\(\displaystyle \begin{array}{rcl}
\frac{{1 - \cos (2x)}}{{1 + \cos (2x)}} & = & \frac{{1 - \left[ {1 - 2\sin ^2 (x)} \right]}}{{1 + \left[ {2\cos ^2 (x) - 1} \right]}} \\
& = & \frac{{2\sin ^2 (x)}}{{2\cos ^2 (x)}} \\
\end{array}\)
\(\displaystyle \begin{array}{rcl}
\frac{{1 - \cos (2x)}}{{1 + \cos (2x)}} & = & \frac{{1 - \left[ {1 - 2\sin ^2 (x)} \right]}}{{1 + \left[ {2\cos ^2 (x) - 1} \right]}} \\
& = & \frac{{2\sin ^2 (x)}}{{2\cos ^2 (x)}} \\
\end{array}\)
Re: Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. an
Hello, smsmith!
You're expected to know these identities:
. . \(\displaystyle \sin^2x \:=\:\frac{1\,-\,\cos2x}{2}\;\;\;\;\;\cos^2x \:=\:\frac{1\,+\,\cos2x}{2}\)
We have: \(\displaystyle \:\tan^2x \:=\:\L\left(\frac{\sin x}{\cos x}\right)^2\:=\:\frac{\sin^2x}{\cos^2x} \:=\:\frac{\frac{1\,-\,\cos2x}{2}}{\frac{1\,+\,\cos2x}{2}} \:=\:\frac{1\,-\,\cos2x}{1\,+\,\cos2x}\)
Hello, smsmith!
You're expected to know these identities:
. . \(\displaystyle \sin^2x \:=\:\frac{1\,-\,\cos2x}{2}\;\;\;\;\;\cos^2x \:=\:\frac{1\,+\,\cos2x}{2}\)
We have: \(\displaystyle \:\tan^2x \:=\:\L\left(\frac{\sin x}{\cos x}\right)^2\:=\:\frac{\sin^2x}{\cos^2x} \:=\:\frac{\frac{1\,-\,\cos2x}{2}}{\frac{1\,+\,\cos2x}{2}} \:=\:\frac{1\,-\,\cos2x}{1\,+\,\cos2x}\)