Question 237313
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If *[tex \Large \sin\Theta\ =\ \frac{4}{5}], then *[tex \Large \Theta] has to be the larger acute angle in a right triangle with sides in proportion 3:4:5.  Therefore, the magnitude of *[tex \Large \cos\Theta] must be *[tex \Large \frac{3}{5}], and since *[tex \Large \cos] is negative in Quadrant II, *[tex \Large \cos\Theta\ =\ -\frac{3}{5}].


Now use *[tex \Large \cot\Theta\ =\ \frac{\cos\Theta}{\sin\Theta}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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