# How Do You Derive the Formula for sin(x-y)?

• B
• farfromdaijoubu
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)##?

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

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

SammyS
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.

FactChecker said:
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.

Last edited:
FactChecker
Bosko said:
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.

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