# [SOLVED] Having trouble with this identity

#### Lucian

I've been at this for a little while now but I'm not quite sure how to proceed. Would someone please help me?

Cos3(theta) = 4Cos^3(theta) - 3Cos(theta)

#### Lucian

Also I'm only supposed to manipulate a single side of the equation.

#### skeeter

MHF Helper
I've been at this for a little while now but I'm not quite sure how to proceed. Would someone please help me?

Cos3(theta) = 4Cos^3(theta) - 3Cos(theta)
$$\displaystyle \cos(3t) =$$

$$\displaystyle \cos(2t+t) =$$

$$\displaystyle \cos(2t)\cos(t) - \sin(2t)\sin(t) =$$

$$\displaystyle [2\cos^2(t) - 1]\cos(t) - 2\sin^2(t)\cos(t) =$$

$$\displaystyle [2\cos^2(t) - 1]\cos(t) - 2[1 - \cos^2(t)]\cos(t) =$$

finish it ...

Lucian

#### Lucian

I think I've got it, thank you for the help Skeeter!

(2Cos^3(t)-Cos(t)) - (2)(Cos(t)-Cos^3(t))
(2Cos^3(t)-Cos(t)) - (2Cos(t) - 2Cos^3(t))
2Cos^3(t) -Cos(t) - 2Cos(t) + 2Cos^3(t)
4Cos^3(t) - 3Cos(t)