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\( \begin{array}{llll}\text { (c) } \sec \left(\frac{23 \pi}{12}\right) & \text { (f) } \csc \left(\frac{7 \pi}{4}\right) & \text { (g) } \cot \left(\frac{3 \pi}{4}\right) & \text { (h) } \sin \left(\frac{19 \pi}{12}\right) \\ \text { (i) } \tan \left(-\frac{11 \pi}{6}\right) & \text { (j) } \cot \left(\frac{13 \pi}{2}\right) & \text { (k) } \csc \left(-\frac{7 \pi}{4}\right) & \text { (l) } \cot (225)\end{array} \)

Ask by Ruiz Huang. in South Africa
Feb 22,2025

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Here are the simplified results for each trigonometric function: - (c) \( \sec \left(\frac{23 \pi}{12}\right) = \frac{4}{\sqrt{6} + \sqrt{2}} \) - (f) \( \csc \left(\frac{7 \pi}{4}\right) = -\sqrt{2} \) - (g) \( \cot \left(\frac{3 \pi}{4}\right) = -1 \) - (h) \( \sin \left(\frac{19 \pi}{12}\right) = -\frac{\sqrt{6} + \sqrt{2}}{4} \) - (i) \( \tan \left(-\frac{11 \pi}{6}\right) = \frac{1}{\sqrt{3}} \) - (j) \( \cot \left(\frac{13 \pi}{2}\right) = \text{undefined} \) - (k) \( \csc \left(-\frac{7 \pi}{4}\right) = \sqrt{2} \) - (l) \( \cot(225^{\circ}) = 1 \)

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Extra Insights

Did you know that angles can extend beyond the basic \(0\) to \(2\pi\) range? For instance, \(\frac{23\pi}{12}\) is equivalent to a larger angle that can be worked into the standard unit circle framework by subtracting \(2\pi\) (or \(12\pi/12\)). So, \(\sec\left(\frac{23\pi}{12}\right)\) actually falls into the fourth quadrant, where the secant function is negative. When evaluating trigonometric functions like \(\csc\left(-\frac{7\pi}{4}\right)\), keep in mind that the cosecant is the reciprocal of the sine. The angle \(-\frac{7\pi}{4}\) is equivalent to \(\frac{\pi}{4}\) in the positive direction (as it wraps around the unit circle), which means \(\csc\left(-\frac{7\pi}{4}\right) = \csc\left(\frac{\pi}{4}\right) = \frac{1}{\sin\left(\frac{\pi}{4}\right)} = \sqrt{2}\).

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