Question 586488
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Secant is just the reciprocal of cosine, so if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec\left(\theta\right)\ =\ -\frac{13}{12}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\left(\theta\right)\ =\ -\frac{12}{13}]


which is just fine because cosine is negative in Quadrant II.


Then use the Pythagorean identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(\varphi)\ +\ \sin^2(\varphi)\ =\ 1]


from which we can derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(\varphi)\ =\ \pm\sqrt{1\ -\ \cos^2(\varphi)}]


And you choose *[tex \LARGE \pm] based on the quadrant.


For you, sine is positive in QII, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\theta\right)\ =\ \sqrt{1\ -\ \frac{144}{169}}\ =\ \sqrt{\frac{25}{169}}\ =\ \frac{5}{13}]


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