Trigonometric formulae

Moubinool

$\sin(ax)\sin(bx)\sin(cx)=\frac{-1}{4}\Letf( -\sin{(a-b+c)x}}-\sin{(-a+b+c)x}-\sin{(a+b-c)x}} +\sin{(a+b+c)x} \Right)$

$\sin(ax)\cos(bx)\cos(cx)=\frac{-1}{4}\Letf( -\sin{(a+b+c)x}}-\sin{(-a+b+c)x}}-\sin{(a+b-c)x}+\sin{(a-b+c)x} \Right)$

$\cos(ax)\sin(bx)\sin(cx)=\frac{1}{4}\Letf( \cos{(a+b-c)x}}+\cos{(a-b+c)x}}-\cos{(a+b+c)x}-\cos{(-a+b+c)x} \Right)$

$\cos(ax)\cos(bx)\cos(cx)=\frac{1}{4}\Letf( \cos{(a+b+c)x}+\cos{(-a+b+c)x}}+\cos{(a-b+c)x}}+\cos{(a+b-c)x} \Right)$

**Posted:** Sun Jan 09, 2005 11:17 am

Kent Merryfield

If asked to prove these, I would immediately replace a cosines and sines by their complex equivalents. That is,

$\cos(x)=\frac{e^{ix}+e^{-ix}}2}$

$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

It's the only way I know to keep from going crazy trying to keep something like this organized.

**Posted:** Sun Jan 09, 2005 11:51 am

Moubinool

In fact it is enough to prove the first one with complex numbers as
Kent said.

For the others use derivation according to $a\; b\; or \; c$

**Posted:** Sun Jan 09, 2005 12:17 pm