Question 1118435
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*[illustration EqOfTerminalRay.jpg].


The slope of the line is the tangent of the angle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \tan\theta\ =\ -5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{\sin\theta}{\cos\theta}\ =\ -5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin\theta\ =\ -5\cos\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin^2\theta\ =\ 25\cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  1\ -\ \cos^2\theta\ =\ 25\cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  26\cos^2\theta\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos^2\theta\ =\ \frac{1}{26}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos\theta\ =\ \pm\frac{\sqrt{26}}{26}]


But the angle is in QII where cosine is negative, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos\theta\ -\frac{\sqrt{26}}{26}]


Make the substitution *[tex \LARGE \cos^2\theta\ =\ 1\ -\ \sin^2\theta] to solve for *[tex \LARGE \sin\theta].  Remember sine is positive in QII.


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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