# Trigonometric identity from Euler's intro to analysis of infinite

1. May 2, 2013

### EvenSteven

So I'm trying to get through euler's introduction to the analysis of the infinite so I could eventually read his books on calculus but I'm stuck somewhere and can't seem to figure out how he equates this identity

so by expanding I get sin(2y) * cos(z) + cos(2y) * sin(z).

I get that the second term of the equation [cos(2y) * sin(z)] gives cos2y - sin2y which gives - sin z if you take the negative out of the factor but I can't seem to figure out how sin(2y) * cos(z) = 2cosy * sin(y + z). I just get 2siny * cosy * cosz and can't seem to do anything else.

I'm probably missing something terribly obvious like always happens but I'd really appreciate some help I have nobody I can ask in real life and I don't like to just move on because the book gets more complicated. This is on page 104 if anyone cares to know.

Last edited: May 2, 2013
2. May 2, 2013

### LCKurtz

Left side:$$\sin(2y+z) = \sin(2y)\cos z + \cos(2y)\sin z = 2\sin y\cos y \cos z + (2\cos^2y - 1)\sin z$$
Right side:$$2\cos y(\sin y \cos z + \cos y \sin z)-\sin z=2\cos y \sin y \cos z +2\cos^2 y \sin z -\sin z$$Does that do it for you?

3. May 2, 2013

### EvenSteven

Yes, thank you very much, figures I just had to use basic identify $$sin^2x = 1 - cos^2x$$ and then simply factor to get to euler's answer.

I definitely should have seen that.