having problems on these trig questions, not really sure how to go about them. If you could tell me the method to go about it would be great, thanks!
1
a) Find $\theta$ such that $\sin(\theta) = \sin(99\pi/5) \quad \text{and} \quad -\frac {1}2 \pi \leq \theta \leq \frac 12 \pi$
b) Find $\theta$ such that $\cos(\theta) = \cos(-94\pi/7) \quad \text{and} \quad 0\pi \leq \theta \leq \pi$
2
Suppose $x$ is in the third quadrant, and $\sin x = -1/3$. Find, without using any of the trig capabilities of your calculator, each of the following:
a) $\cos x$
b) $\sin 2x$
c) $\cos 2x$
d) $\sin\left(\dfrac{x}{2}\right)$
Best Answer
For question #1 you need to use the formula: $$\sin \theta = \sin a\implies \theta =\begin {cases} 2 k \pi + a,& k \in \mathbb Z \\\\ \text{or}\\\\ 2k\pi +\pi - a,&k \in \mathbb Z \end{cases}$$ and in the second case $$\cos \theta = \cos a\implies \theta = 2k\pi \pm a, \quad k \in \mathbb Z$$
In our example, in the first case $a = 99\pi/5.$ Take advantage of the inequality to find all the appropriate $k\in \mathbb Z$ in order to define $\theta$. The concept is the same for the other part of the same question.
For question #2 you need to take advantage of some fundamental trigonometric identities, e.g. $$\sin^2 x=\frac{1-\cos 2x}{2}$$