Question 973152
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You have described an impossible situation.  Sine is negative in the third quadrant.  However, I'm going to proceed assuming the highly likely event that you meant to type *[tex \Large \sin\theta\ =\ -\frac{15}{17}].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ -\frac{15}{17}]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ \frac{225}{289}]


Use the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\theta\ =\ \frac{225}{289}]


Algebra and arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ =\ \frac{64}{289}]


Take the square root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\frac{8}{17}]


But cosine is negative in QIII, so discard the positive root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ -\frac{8}{17}]


Of course all of this is valid ONLY to the extent that my assumption about your typing error was correct.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \