+0

# pls help

0
61
1

i am really confused

Apr 17, 2020

#1
+24933
+3

pls help
$$\text{\cos(\theta) = -\dfrac{\sqrt{2}}{3}, where \pi \leq \theta\leq \dfrac{3\pi}{2}. [quadrant III] }$$
$$\text{\tan(\beta) = \dfrac{4}{3}, where 0 \leq \beta \leq \dfrac{\pi}{2}. [quadrant I] }$$
What is the exact value of $$\sin(\theta+\beta)$$

$$\begin{array}{rcll} \cos(\theta_{III}) &=& -\dfrac{\sqrt{2}}{3} \\ \tan(\beta_{I}) &=& \dfrac{4}{3} \\ \end{array}$$

$$\begin{array}{|rcll|} \hline \sin(\theta_{III}) &=& - \sqrt{1-\cos^2(\theta_{III})} \quad | \quad \boxed{\sin(\theta_{III}) < 0\ !} \\ \sin(\theta_{III}) &=& - \sqrt{1- \left(-\dfrac{\sqrt{2}}{3} \right)^2 } \\ \sin(\theta_{III}) &=& - \sqrt{1- \dfrac{2}{9} } \\ \sin(\theta_{III}) &=& - \sqrt{\dfrac{9-2}{9} } \\ \sin(\theta_{III}) &=& - \sqrt{\dfrac{7}{9} } \\ \mathbf{\sin(\theta_{III})} &=& \mathbf{- \dfrac{\sqrt{7}}{3}} \\ \hline \end{array}$$

Formula:

$$\begin{array}{|rcll|} \hline \sin^2(\beta)+\cos^2(\beta) &=& 1 \quad | \quad : \cos^2(\beta) \\ \tan^2(\beta)+1 &=& \dfrac{1}{\cos^2(\beta)} \\ \cos^2(\beta) &=& \dfrac{1}{1+\tan^2(\beta)} \\ \mathbf{\cos(\beta)} &=& \mathbf{\pm\sqrt{ \dfrac{1}{1+\tan^2(\beta)} } } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \cos(\beta_{I}) &=& +\sqrt{ \dfrac{1}{1+\tan^2(\beta_{I})} } \quad | \quad \boxed{\cos(\beta_{I}) > 0\ !} \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{1+ \left(\dfrac{4}{3}\right)^2} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{1+ \dfrac{16}{9}} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{\dfrac{9+16}{9}} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{1}{\dfrac{25}{9}} } \\ \cos(\beta_{I}) &=& \sqrt{ \dfrac{9}{25} } \\ \mathbf{\cos(\beta_{I})} &=& \mathbf{\dfrac{3}{5}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \sin(\beta_{I}) &=& \tan(\beta_{I})\cos(\beta_{I}) \\ \sin(\beta_{I}) &=& \left(\dfrac{4}{3}\right)\left(\dfrac{3}{5}\right) \\ \mathbf{\sin(\beta_{I})} &=& \mathbf{\dfrac{4}{5}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\sin(\theta_{III}+\beta_{I}) } &=& \mathbf{\sin(\theta_{III})\cos(\beta_{I})+\cos(\theta_{III})\sin(\beta_{I}) } \\ \\ \sin(\theta_{III}+\beta_{I}) &=& \left( - \dfrac{\sqrt{7}}{3} \right) \left( \dfrac{3}{5} \right) + \left( -\dfrac{\sqrt{2}}{3}\right) \left( \dfrac{4}{5}\right) \\ \sin(\theta_{III}+\beta_{I}) &=& -\dfrac{3\sqrt{7}}{15} -\dfrac{4\sqrt{2}}{15} \\ \mathbf{\sin(\theta_{III}+\beta_{I})} &=& \mathbf{-\dfrac{\sqrt{7}}{5} -\dfrac{4\sqrt{2}}{15}} \\ \hline \end{array}$$

Apr 17, 2020