Deriving formula for sin(x-y)

[farfromdaijoubu]

TL;DR Summary

In trying to derive the compound angle formula for sin(x-y), I ended up with two possible solutions from the quadratic. How do I know which solution to take?

I was trying to show that $sin(x-y) = sin(x)cos(y)-cos(x)sin(y)$ using Pythagoras' theorem and $cos(x-y)=cos(x)cos(y)+sin(x)sin(y)$.

I have:
$$sin^2(x-y)=1-cos^2(x-y)$$
$$sin^2(x-y)=1-(cos(x)cos(y)+sin(x)sin(y))^2$$
$$sin^2(x-y)=1-(cos^2(x)cos^2(y)+sin^2(x)sin^2(y)+2cos(x)cos(y)sin(x)sin(y))$$
$$sin^2(x-y)=1-(cos^2(x)(1-sin^2(y))+sin^2(x)(1-cos^2(y))+2cos(x)cos(y)sin(x)sin(y))$$
$$sin^2(x-y)=(cos^2(x)sin^2(y)+sin^2(x)cos^2(y)-2cos(x)cos(y)sin(x)sin(y))$$
$$sin^2(x-y)=(cos(x)sin(y)-sin(x)cos(y))^2$$

But now how do you know if it's $sin(x-y)=cos(x)sin(y)-sin(x)cos(y)$ or $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$?

 

[anuttarasammyak]

For y=0
The first one says
$$\sin x=-\sin x$$
The senond one says
$$\sin x=\sin x$$

 

[PeroK]

You could note that$$\sin(x-y) = \cos(x-y -\frac{\pi}2)$$Then apply the cosine result to the right-hand side.

 

[FactChecker]

FYI, you will be able to solve these problems much more methodically if you are familiar with Euler's formula, $e^{ix}=\cos(x)+i\sin(x)$, and the associated identities, $\sin(x) = (e^{ix}-e^{-ix})/2i$ and $\cos(x) = (e^{ix}+e^{-ix})/2$.

PS. I can not see your profile, so I do not know if your mathematical background should have already covered these identities.

 

[Bosko]

FYI, you will be able to solve these problems much more methodically if you are familiar with Euler's formula, $e^{ix}=\cos(x)+i\sin(x)$, and the associated identities, $\sin(x) = (e^{ix}-e^{-ix})/2i$ and $\cos(x) = (e^{ix}+e^{-ix})/2$.

PS. I can not see your profile, so I do not know if your mathematical background should have already covered these identities.

There are more direct way. I am unable to remember those formulae and I'm using the following way to prove/remember pairs of them, each time I need them.
$$\begin{align}
\cos(x-y)+i\sin(x-y)&=e^{i(x-y)}\nonumber \\
&=e^{ix-iy}\nonumber \\
&=e^{ix} \cdot e^{-iy}\nonumber \\
&=e^{ix} \cdot e^{i(-y)}\nonumber \\
&=\left[\cos(x)+i\sin(x)\right]\cdot\left[\cos(-y)+i\sin(-y)\right]\nonumber \\
\end{align}$$
Using parity of sine and cosine functions and multiplication at the end, pairs of the required formulas are obtained.

 

[FactChecker]

There are more direct way. I am unable to remember those formulae and I'm using the following way to prove/remember pairs of them, each time I need them.
$$\begin{align}
\cos(x-y)+i\sin(x-y)&=e^{i(x-y)}\nonumber \\
&=e^{ix-iy}\nonumber \\
&=e^{ix} \cdot e^{-iy}\nonumber \\
&=e^{ix} \cdot e^{i(-y)}\nonumber \\
&=\left[\cos(x)+i\sin(x)\right]\cdot\left[\cos(-y)+i\sin(-y)\right]\nonumber \\
\end{align}$$
Using the parity of sine and cosine and multiplication at the end, pairs of the required formulas are obtained.

Very good. As I wrote my post, I kept thinking that it there was a better way but I couldn't remember it.