Originally Posted by

**sadmath** I need help understanding these trig identities:

$\displaystyle sin(a)cos(b) = (sin(a+b) + sin(a-b))/2$

Where I see this identity I also see it followed up by this identity:

$\displaystyle cos(b)sin(a) = (sin(b+a) - sin(b-a))/2$

Huh? I recall that:

$\displaystyle sin(-u) = -sin(u)$

So cos(b)sin(a) becomes:

$\displaystyle cos(b)sin(a) = (sin(b+a) + sin(a-b))/2$

Right? but this is NOT equal to the sin(a)cos(b) equation above. Shouldn't it be?

I know that xy=yx so shouldn't cos(b)sin(a) = sin(a)cos(b)?

Where is my logic flawed...