Author Topic: TT1 Problem 1 (night)  (Read 753 times)

Victor Ivrii

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TT1 Problem 1 (night)
« on: October 19, 2018, 04:12:07 AM »
Show that
\begin{equation*}
|\sin (z)|^2 = \sin^2 (x)+ \sinh^2 (y)
\end{equation*}
 for all complex numbers $z = x+yi$.

ZhenDi Pan

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Re: TT1 Problem 1 (night)
« Reply #1 on: October 19, 2018, 05:55:03 AM »
My solution:
\begin{align*}
| \sin(z)^{2}| & =[\sin (x) \cos(iy)]^{2}+[\cos (x)\sin(iy)]^{2}\\
 & = [\sin^{2}(x)\cosh^{2}(y)]+[\cos^{2}(x)\sinh^{2}(y)] \\
 & = [\sin^2(x)(1+\sinh^2(y))]+[(1-\sin^2(x))\sinh^2(y)] \\
 & = \sin^2(x)+\sinh^2(y)-\sin^2(x)\sinh^2(y)+\sin^2(x)\sinh^2(y) \\
 & = \sin^2(x)+\sinh^2(y)
\end{align*}
Please reformat (see how I changed your first line)
« Last Edit: October 19, 2018, 06:21:07 AM by ZhenDi Pan »