Toronto Math Forum
MAT3342018F => MAT334Tests => Term Test 1 => Topic started by: Victor Ivrii 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 $z = x+yi$.

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)