# Expanding sin(x-iy)

I’m self studying through Stroud’s Engineering Mathematics 7th Ed. and can’t seem to figure out how one of the answers was arrived at.

Here is the question:

“Obtain the expansion of $$\sin(x-iy)$$ in terms of the trigonometric and hyperbolic functions of $$x$$ and $$y$$.”

Using trig identities I get:

$$\sin(x)\cos(iy) - \cos(x)\sin(iy)$$

Then I use $$\cos(iy) = \cosh(y)$$, $$\sin(iy) = i\sinh(y)$$ and $$\cos(x) = \cosh(ix)$$

Which gives me:

$$\sin(x)\cosh(y) - \cosh(ix)i\sinh(y)$$

However, the solution provided is $$\sin(x)\cosh(y) - i\cos(x)\sinh(y)$$

I’ve looked up errata in case it’s incorrect, but it’s not mentioned in the errata, so I suspect I’ve made an error somewhere.

Hoping someone can shed some light.

Thanks.

• You were asked for trig & hyperbolic functions of $x$ and $y$; you gave an answer that included a hyperbolic function of $ix$. Commented Feb 24, 2020 at 3:59
• Thanks, Gerry. This actually helped in two ways: you pointed out that I absent-mindedly converted cosx to cosh(ix) when I shouldn’t have, and then I realised why may answer looked different when I had played with the cosx notion previously. I had tunnel vision expecting the sin(iy) to transform to isinh(y) and couldn’t figure out how the cosx in the answer had transformed to icosx, but after stepping back and looking at it now, it’s clear the “i” comes from the isinh(y) as it has simply been moved to the front. Thanks! Commented Feb 24, 2020 at 8:39

By definition, $$\sin(z)=\frac{i}{2}(e^{-iz}-e^{iz})$$ Thus $$\sin(x-iy)=\frac{i}{2}(e^{-y-ix}-e^{y+ix})$$ $$=\frac{i}{2}(e^{-y}(\cos(x)-i\sin(x))-e^y(\cos(x)+i\sin(x)))$$ You can take it from there.