Question

# 4. (3 points) Suppose that a is an acute angle (measured in radians). Use trigonometric identities, co-function identities, or reference angles to rewrite each of the trigonometric function values below

in terms of sin(a) or cos() (or both) only. In some cases, it might be helpful to make a little sketch of the situation,either on paper or in your mind, to help you solving the exercises. To illustrate what I expect you to do, the first two parts are already solved, so you need to start the exercise by solving (c). \begin{aligned} &\text { (a) } \sec \left(\frac{\pi}{2}-\alpha\right) \text { . Solution. We know that } \sec \left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\cos \left(\frac{\pi}{2}-\alpha\right)} . \text { Since sine and cosine are cofunc- }\\ &\text { tions, we also know that } \cos \left(\frac{\pi}{2}-\alpha\right)=\sin (\alpha) . \text { Therefore, } \sec \left(\frac{\pi}{2}-\alpha\right)=\frac{1}{\sin (\alpha)} \text { . } \end{aligned} \text { (c) } \cot \left(\frac{\pi}{2}-\alpha\right) \text { . } \text { (d) } \tan (\pi+\alpha) \text { . } \text { (b) } \cos (\pi-\alpha) \text { . Solution. The angle } \beta=\pi-\alpha \text { is on the second quadrant, and its reference angle is } \alpha \text { . } \text { Thus, } \cos (\pi-\alpha)=-\cos (\alpha) \text { (the negative sign is because of } \pi-\alpha \text { being on the second quadrant). } \text { (e) } \tan (\pi-\alpha) \text { . } \text { (g) } \cos (2 \pi+\alpha) \text { (f) } \sin (2 \pi-\alpha) \text { . } \text { (h) } \cos (\alpha-2 \pi) \text { . }

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