- #1

farfromdaijoubu

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- 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)##?

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)##?