Another one from my assignment:

Assume $$ \tan (\frac{x}{2})=\tan A\tanh B $$

Prove that $$ \tan(x)=\frac{\sin(2A)\sinh(2B)}{1+cos(2A)\cosh(2B)} $$

I manipulated $$ \tan(x)=\tan2(\frac {x}{2})=\frac {2\tan A \tanh B}{1-(\tan A\tanh B)^2} $$ Using the half angle formula for tan. From there I seem to just be going in circles and coming back to the original question every attempt.


We will use the formulas in this page. Then $$\sin(2A)=2\sin(A)\cos(A)=\frac{2\tan(A)}{1/\cos^2(A)}=\frac{2\tan(A)}{1+\tan^2(A)}.$$ In a similar way we can express $\cos(2A)$, $\sinh(2B)$, $\cosh(2B)$ in terms of $\tan(A)$ or $\tanh(B)$.

Hence $$\tan(x)=\frac{\sin(2A)\sinh(2B)}{1+\cos(2A)\cosh(2B)}\\=\frac{2\tan A}{1+\tan^2 A}\cdot \frac{2\tanh B}{1-\tanh^2 B} \cdot \left(1+\frac{1-\tan^2 A}{1+\tan^2 A}\cdot\frac{1+\tanh^2 B}{1-\tanh^2 B}\right)^{-1}\\= \frac{2\tan A \tanh B}{1-(\tan A\tanh B)^2}.$$

On the other hand $$\tan(x)=\tan 2(x/2)=\frac{2\tan A \tanh B}{1-(\tan A\tanh B)^2}.$$

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  • $\begingroup$ Hi, I asked my lecturer about it now. He only changed one thing in the question, changing $$ \tan(x)=\frac{\sin(2A)sinh(2B)}{\cos (2A)cosh (2B)} $$ to $$ tan (x)=\frac{sin (2A)sinh(2B)}{1+\cos(2A)cosh(2B)} $$. Don't know if that changes the logic much? $\endgroup$ – Rene Aug 10 '16 at 8:23
  • $\begingroup$ @Rene The logic is similar. Take a look to my revised answer. $\endgroup$ – Robert Z Aug 10 '16 at 9:10

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