Trig. Identities (cos3x)

**Savior_Self** · April 18th 2010, 05:51 PM

Alright, I'll write what I have so far..

Show: $\cos3x = 4\cos^3x - 3\cos x$ (changing only $\cos3x$ )

---

considering that $\cos3x = \cos(2x + x)$ and $\cos(A+B) = \cos A\cos B - \sin A\sin B$

we get...
$\cos2x\cos x - \sin2x\sin x$

using the double angle identities...

$(\cos^2 x - \sin^2 x)\cos x - (2\sin x\cos x)\sin x$

$\cos^3 x - (\sin^2 x)\cos x - 2\sin^2 x\cos x$

Using the Pythagorean identity.. $2\sin^2 x = 2(1 -\cos^2 x)$

$\cos^3 x - (1 - \cos^2 x)\cos x - (2 - 2\cos x)\cos x$

$\cos^3 x - (\cos x - \cos^3 x) - 2\cos x + 2\cos^2 x$

$\cos^3 x - \cos x + \cos^3x - 2\cos x + 2cos^2 x$

$2\cos^3x - 3\cos x + 2\cos^2x$

So I don't really know where to go from here... and I'm sure I made mistakes thus far. So could anyone help me out here? I'd really appreciate it. Thanks guys.

**skeeter** · April 18th 2010, 06:12 PM

Originally Posted by Savior_Self

Alright, I'll write what I have so far..

Show: $\cos3x = 4\cos^3x - 3\cos x$ (changing only $\cos3x$ )

---

considering that $\cos3x = \cos(2x + x)$ and $\cos(A+B) = \cos A\cos B - \sin A\sin B$

we get...
$\cos2x\cos x - \sin2x\sin x$

using the double angle identities...

$(\cos^2 x - \sin^2 x)\cos x - (2\sin x\cos x)\sin x$

$\cos^3 x - (\sin^2 x)\cos x - 2\sin^2 x\cos x$

Using the Pythagorean identity.. $2\sin^2 x = 2(1 -\cos^2 x)$

$\cos^3 x - (1 - \cos^2 x)\cos x - (2 - 2\cos\textcolor{red}{^2}{x})\cos x$

...

**Savior_Self** · April 18th 2010, 06:25 PM

hahaha

thanks, man.

**Indian25** · April 19th 2010, 01:49 AM

Cos3x=cos(x+2x) and

cos3x=cosx.cos2x-sinxsin2x
=cosx(2cos^2 x-1)-sinx(2sinxcosx)
=2cos^3x-cosx-2sin^2xcosx
=2cos^3x-cosx-2(1-cos^2x)cosx
=2cos^3x-cosx-2cosx+2cos^3x
=4cos^3x-3cosx

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