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I have $\cos^2x\cosh^2y - \sin^2x\sinh^2y$. I saw it written simplified as $\cosh^2 y - \sin^2 x$. But I don't get how to get it.

My attempts were to write $\cosh^2y -1$ instead of $\sinh^2y$ but that way I get $\cos^2x\cosh^2y - \sin^2x\sinh^2y = \cosh^2y(\cos^2x-\sin^2x) + \sin^2x$.

What am I doing wrong?

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    $\begingroup$ Should it be $$\cos^2x\cosh^2y+\sin^2x\sinh^2y=(1-\sin^2x)\cosh^2y+\sin^2x(\cosh^2y-1)$$ $\endgroup$ – lab bhattacharjee Nov 12 '18 at 10:33
  • $\begingroup$ then I have $\cos^2x\cosh^2y+\sin^2x\sinh^2y=(1-\sin^2x)\cosh^2y-\sin^2x(\cosh^2y-1) = \cosh^2y - \sin^2x\cosh^2 - \sin^2x\cosh^y + \sin^2x = \cosh^2y - 2\sin^2x\cosh^2 + \sin^2x$. Which is not equal to $\cosh^2y-\sin^2x$ $\endgroup$ – user3132457 Nov 12 '18 at 16:04
  • $\begingroup$ seems like there should be + instead of -. In that case I'm getting the right answer. $\endgroup$ – user3132457 Nov 12 '18 at 16:22
  • $\begingroup$ I was actually finding real, imaginary parts of $\tan(x+yi)$ and the modulus of it. I have $$Re(\tan(x+yi)) = {\frac{\sin2x}{\cos2x+\cosh2x}}, Im(\tan(x+yi))= {\frac{\sinh2y}{\cos2x+\cosh2x}}$$. But I can't seem to find the modulus. The answer is $${\sqrt{\frac{\cosh2y-\cos2x}{\cosh2y+\cosh2x}}}$$ $\endgroup$ – user3132457 Nov 12 '18 at 18:50
  • $\begingroup$ Bottom of fraction should be $\cosh2y+\cos2x$ $\endgroup$ – user3132457 Nov 12 '18 at 19:14

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