Question 509195
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Let's look at the unit circle:


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


Note that *[tex \Large -\frac{7\pi}{6}\ =\ \frac{5\pi}{6}]


And recall that *[tex \Large \sec\varphi\ =\ \frac{1}{\cos\varphi}]


So, find *[tex \Large \frac{5\pi}{6}] on the unit circle, note that the *[tex \Large x]-coordinate is *[tex \Large -\frac{\sqrt{3}}{2}], take the reciprocal and rationalize to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec\left(-\frac{7\pi}{6}\right)\ =\ -\frac{2\sqrt{3}}{3}]


Which is what you had.


Now, recall that *[tex \Large \csc\varphi\ =\ \frac{1}{\sin\varphi}]


So, find *[tex \Large \frac{\pi}{3}] on the unit circle, note that the *[tex \Large y]-coordinate is *[tex \Large \frac{\sqrt{3}}{2}], take the reciprocal and rationalize to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\left(\frac{\pi}{3}\right)\ =\ \frac{2\sqrt{3}}{3}]


Hence the sum of your first two terms is zero.


Now for the third term:


Note that *[tex \Large \frac{8\pi}{3}\ =\ \frac{6\pi}{3}\ +\ \frac{2\pi}{3}\ =\ 2\pi\ +\ \frac{2\pi}{3}]


Since the tangent function has a periodicity of *[tex \Large \pi], we can conclude that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\left(\frac{8\pi}{3}\right)\ =\ \tan\left(\frac{2\pi}{3}\right)]


Now, recall that *[tex \Large \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


So, find *[tex \Large \frac{2\pi}{3}] on the unit circle, note that the *[tex \Large y]-coordinate is *[tex \Large \frac{\sqrt{3}}{2}], the *[tex \Large x]-coordinate is *[tex \Large \frac{-1}{2}], and then take the quotient of the *[tex \Large y]-coordinate divided by the *[tex \Large x]-coordinate to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\left(\frac{2\pi}{3}\right)\ =\ \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\ =\ -\sqrt{3}]


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