# If sinalpha=3/5 and alpha lies in Q2 and sinbeta=5/13 and beta lies in Q1, find cos(alpha+beta), cosbeta and tanalpha?

Mar 16, 2017

$\cos \left(\alpha + \beta\right) = - \frac{63}{65}$; $\cos \beta = \frac{12}{13}$ and $\tan \alpha = - \frac{3}{4}$

#### Explanation:

As $\sin \alpha = \frac{3}{5}$ and $\alpha$ lies in Q2, $\cos \alpha$ is negative and

$\cos \alpha = \sqrt{1 - {\sin}^{2} \alpha} = \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}}$

= $\sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = - \frac{4}{5}$

and $\tan \alpha = \sin \frac{\alpha}{\cos} \alpha = \frac{\frac{3}{5}}{- \frac{4}{5}} = \frac{3}{5} \times \left(- \frac{5}{4}\right) = - \frac{3}{4}$

further as $\sin \beta = \frac{5}{13}$ and $\beta$ lies in Q1, $\cos \alpha$ is positive and

$\cos \beta = \sqrt{1 - {\sin}^{2} \beta} = \sqrt{1 - {\left(\frac{5}{13}\right)}^{2}}$

= $= \sqrt{1 - \frac{25}{169}} \sqrt{\frac{144}{25}} = \frac{12}{13}$

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

= $\cos \alpha \cos \beta - \sin \alpha \sin \beta$

= $- \frac{4}{5} \times \frac{12}{13} - \frac{3}{5} \times \frac{5}{13}$

= $- \frac{48}{65} - \frac{15}{65} = - \frac{63}{65}$