1
$\begingroup$

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) $\sin\left(\dfrac{x}{2}\right)$

$\endgroup$
  • 2
    $\begingroup$ Start simplifying the rhs : $\frac{99\pi}{5}=20\pi-\frac \pi 5$ and $\frac{94\pi}{7}=13\pi+\frac{3 \pi}7$ $\endgroup$ – Claude Leibovici Mar 22 '16 at 8:09
0
$\begingroup$

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. $$\sin^2 x=\frac{1-\cos 2x}{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.