# How to solve (sin^2(alpha+beta)-cos^2alpha-cos^2beta)/(sin(alpha+beta)-sin^2alpha-sin^2beta) if alpha=pi/3 and beta=2pi/3?

Jun 12, 2018

Given $\alpha = \frac{\pi}{3}$ and $\beta = \frac{2 \pi}{3}$

We have $\left(\beta + \alpha\right) = \pi \mathmr{and} \left(\beta - \alpha\right) = \frac{\pi}{3}$

To evaluate

$\frac{{\sin}^{2} \left(\alpha + \beta\right) - {\cos}^{2} \alpha - {\cos}^{2} \beta}{\sin \left(\alpha + \beta\right) - {\sin}^{2} \alpha - {\sin}^{2} \beta}$

$= \frac{{\sin}^{2} \left(\alpha + \beta\right) - \left({\cos}^{2} \alpha + {\cos}^{2} \beta\right)}{\sin \left(\alpha + \beta\right) - \left({\sin}^{2} \alpha + {\sin}^{2} \beta\right)}$

$= \frac{{\sin}^{2} \left(\alpha + \beta\right) - \left(1 - {\sin}^{2} \alpha + {\cos}^{2} \beta\right)}{\sin \left(\alpha + \beta\right) - \left({\sin}^{2} \alpha + 1 - {\cos}^{2} \beta\right)}$

$= \frac{{\sin}^{2} \left(\alpha + \beta\right) - \left(1 + {\cos}^{2} \beta - {\sin}^{2} \alpha\right)}{\sin \left(\alpha + \beta\right) - \left(1 - \left({\cos}^{2} \beta - {\sin}^{2} \alpha\right)\right)}$

=(sin^2(alpha+beta)-(1+cos(beta+alpha)cos(beta-alpha)))/(sin(alpha+beta)-(1-cos(beta+alpha)cos(beta-alpha))

=(sin^2(pi)-(1+cos(pi)cos(pi/3)))/(sin(pi)-(1-cos(pi)cos(pi/3))

$= \frac{0 - \left(1 - \frac{1}{2}\right)}{0 - \left(1 + \frac{1}{2}\right)} = \frac{1}{3}$

Formula used

$\cos \left(\beta + \alpha\right) \cos \left(\beta - \alpha\right)$

$= {\cos}^{2} \beta {\cos}^{2} \alpha - {\sin}^{2} \beta {\sin}^{2} \alpha$

$= {\cos}^{2} \beta \left(1 - {\sin}^{2} \alpha\right) - \left(1 - {\cos}^{2} \beta\right) {\sin}^{2} \alpha$

$= {\cos}^{2} \beta - {\sin}^{2} \alpha$

Alternative

$\frac{{\sin}^{2} \left(\alpha + \beta\right) - {\cos}^{2} \alpha - {\cos}^{2} \beta}{\sin \left(\alpha + \beta\right) - {\sin}^{2} \alpha - {\sin}^{2} \beta}$

$= \frac{{\sin}^{2} \left(\pi\right) - {\cos}^{2} \left(\frac{\pi}{3}\right) - {\cos}^{2} \left(\frac{2 \pi}{3}\right)}{\sin \left(\pi\right) - {\sin}^{2} \left(\frac{\pi}{3}\right) - {\sin}^{2} \left(\frac{2 \pi}{3}\right)}$

$= \frac{0 - \frac{1}{4} - \frac{1}{4}}{0 - \frac{3}{4} - \frac{3}{4}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3}$