# How do you verify (sin^3x - cos^3x) / (sinx - cosx) = (1 + sin2x) / 2?

Nov 2, 2015

Actually, that should be $\textcolor{b l u e}{\frac{2 + \sin 2 x}{2}}$ or $\textcolor{red}{1 + \frac{\sin 2 x}{2}}$.

#### Explanation:

$\left[1\right] \text{ } \frac{{\sin}^{3} x - {\cos}^{3} x}{\sin x - \cos x}$

Difference of two cubes: ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

$\left[2\right] \text{ } = \frac{\left(\sin x - \cos x\right) \left({\sin}^{2} x + \sin x \cos x + {\cos}^{2} x\right)}{\sin x - \cos x}$

Cancel $\left(\sin x - \cos x\right)$

$\left[3\right] \text{ } = {\sin}^{2} x + \sin x \cos x + {\cos}^{2} x$

Pythagorean Identity: ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\left[4\right] \text{ } = 1 + \sin x \cos x$

Double Angle Identity: $\sin \theta \cos \theta = \frac{1}{2} \sin 2 \theta$

$\left[5\right] \text{ } = 1 + \frac{1}{2} \sin 2 x$

Multiply $1$ by $\frac{2}{2}$ so the two terms will have the same denominator.

$\left[6\right] \text{ } = \frac{2}{2} + \frac{1}{2} \sin 2 x$

$\left[7\right] \text{ } = \frac{2 + \sin 2 x}{2}$

$\textcolor{m a \ge n t a}{\text{ } \therefore \frac{{\sin}^{3} x - {\cos}^{3} x}{\sin x - \cos x} = \frac{2 + \sin 2 x}{2} = 1 + \frac{\sin 2 x}{2}}$