show that the complex version of the addition law for the sine still holds:


We will use the complex definition of $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ and $\cos(z)=\frac{e^{iz}+e^{-iz}}{2i}$ along with defining $z=x+iy$ and $w=a+bi$. The LHS yields $$\frac{e^{i(z+w)}-e^{-i(z+w)}}{2i} $$ $$\frac{e^{iz}e^{iw}-e^{-iz}e^{-iw}}{2i}$$

Substituting in our defined complex values for $z$ and $w$ and Using Euler's identity:


Finding a common denominator:


At this point I got lost so I attempt to use the prior definitions for the RHS to get $$\frac{\frac{[\cos(x)+i\sin(x)]-e^{2y}}{e^y[\cos(x)+i\sin(x)]}}{2i}\cdot\frac{\frac{[\cos(a)+i\sin(a)]+e^{2b}}{e^b[\cos(a)+i\sin(a)]}}{2}+\frac{\frac{[\cos(x)+i\sin(x)]+e^{2y}}{e^y[\cos(x)+i\sin(x)]}}{2}\cdot\frac{\frac{[\cos(a)+i\sin(a)]-e^{2b}}{e^b[\cos(a)+i\sin(a)]}}{2i} $$

And I'm more confident in this result for the RHS than I am for the result I got for the LHS. I'm a bit lost as to where to take this problem from here.


1 Answer 1


Just try to finish the job by exponential: \begin{align*} &\sin z\cos w+\cos z\sin w\\ &=\dfrac{1}{2i}\dfrac{1}{2}\left[(e^{iz}-e^{-iz})(e^{iw}+e^{-iw})+(e^{iz}+e^{-iz})(e^{iw}-e^{-iw})\right]\\ &=\dfrac{1}{2i}\dfrac{1}{2}\left[2e^{i(z+w)}-2e^{-i(z+w)}\right]\\ &=\sin(z+w). \end{align*}

  • 1
    $\begingroup$ Wow, I feel like I was trying to use a chainsaw to cut paper, this is much clearer, thanks. $\endgroup$
    – Peetrius
    Commented Mar 28, 2018 at 3:08

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