Question 1052903
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos^2(\alpha)\ =\ \frac{36}{49}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  1\ -\ \sin^2(\alpha)\ =\ \frac{36}{49}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin^2(\alpha)\ =\ \frac{13}{49}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(\alpha)\ =\ \pm\frac{\sqrt{13}}{7}]


But *[tex \Large \sin] is positive in QII, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(\alpha)\ =\ \frac{\sqrt{13}}{7}]


Similarly


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(\beta)\ =\ -\frac{\sqrt{17}}{9}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos(\alpha\ +\ \beta)\ =\ \cos(\alpha)\cos(\beta)\ -\ \sin(\alpha)\sin(\beta)]


So:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos(\alpha\ +\ \beta)\ =\ \left(-\frac{6}{7}\right)\left(\frac{8}{9}\right)\ -\ \left(\frac{\sqrt{13}}{7}\right)\left(-\frac{\sqrt{17}}{9}\right)]


I'll leave it to you to simplify.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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