# What is cos^3theta-cos^2theta+costheta in terms of non-exponential trigonometric functions?

Feb 17, 2016

${\cos}^{3} \theta - {\cos}^{2} \theta + \cos \theta = \frac{\cos 3 \theta}{4} - \frac{\cos 2 \theta}{2} + \frac{7 \cos \theta}{4} - \frac{1}{2}$

#### Explanation:

Do you know the compound angle identities for $\cos \theta$?

$\cos 2 \theta \equiv 2 {\cos}^{2} \theta - 1$

$\cos 3 \theta \equiv 4 {\cos}^{3} \theta - 3 \cos \theta$

Change the subject of the formula.

${\cos}^{2} \theta \equiv \frac{\cos 2 \theta + 1}{2}$

${\cos}^{3} \theta \equiv \frac{\cos 3 \theta + 3 \cos \theta}{4}$

So substitute them into your expression.

${\cos}^{3} \theta - {\cos}^{2} \theta + \cos \theta$

$= \frac{\cos 3 \theta + 3 \cos \theta}{4} - \frac{\cos 2 \theta + 1}{2} + \cos \theta$

$= \frac{\cos 3 \theta}{4} - \frac{\cos 2 \theta}{2} + \frac{7 \cos \theta}{4} - \frac{1}{2}$

Feb 17, 2016

$f \left(x\right) = \cos x \left(\frac{1 + \cos 2 x}{2} - \cos x + 1\right)$

#### Explanation:

Put cos x in common factor -->
$f \left(x\right) = \cos x \left({\cos}^{2} x - \cos x + 1\right)$.
Replace ${\cos}^{2} x$ by $\left(\frac{1 + \cos 2 x}{2}\right)$
$f \left(x\right) = \cos x \left[\frac{1 - \cos 2 x}{2} - \cos x + 1\right] =$