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) $\sin\left(\dfrac{x}{2}\right)$
$\endgroup$ 13 Answers
$\begingroup$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. $$\sin^2 x=\frac{1-\cos 2x}{2}$$
$\endgroup$ $\begingroup$For 1a), you start at $(1,0)$ and travel a distance $99\pi/5$ counterclockwise around the unit circle, and stop exactly then. Where do you stop? How far around the circle is it from the start if you just go from $(1,0)$ to your stopping point without passing it?
Now the question is asking for an angle in the first or fourth quadrant, so if you stopped in the second or third quadrant you need to reflect the angle across the $y$ axis so that you have something in the first or fourth quadrants at the same distance above or below the $x$ axis; that is how the angle's sine is the same.
Part 1b) uses some similar ideas, but since the angle is negative you go clockwise, and since you are trying to get an equal cosine you need the final angle to give a point at the same distance left or right of the $y$ axis.
For Part 2a) you can use the identity $\sin^2\theta + \cos^2\theta = 1.$ Whenever you know $\sin\theta$ you can use this identity to solve for $\cos^2\theta.$ To get $\cos\theta,$ the square root of $\cos^2\theta$ tells you the magnitude, and by considering which quadrant the angle is in you can deduce the sign.
Once you have solved 2a) you have both $\sin\theta$ and $\cos\theta,$ and there are standard trig formulas for all the remaining questions in Part 2 that just require you to know one or both of those function values.
$\endgroup$ $\begingroup$(1) Write this as $\sin \theta-\sin(99π/5)=0,$ and factor LHS to give $$2\cos\frac12\left(\theta+99π/5\right)\sin\frac12\left(\theta-99π/5\right)=0.$$ Thus you have two possible equations of the form $\cos a=0$ and $\sin b=0.$ These have the solutions $a=(1+2j)\fracπ2$ and $b=πk$ respectively, where $j,k$ are arbitrary integers. Now impose the inequality condition to deduce the specific solutions you need.
(2) Basically the same procedure as above. Here, your equation $\cos\theta-\cos(94π/7)=0$ becomes $$-2\sin\frac12\left(\theta+94π/7\right)\sin\frac12\left(\theta-94π/7\right)=0.$$ Proceed as before.
(3) (a) Use the identity $\sin^2x+\cos^2x=1,$ with the condition that $\cos x$ is also negative. (b) Recall that $\sin 2x=2\sin x\cos x$ and use the above two results. (c) Recall that $\cos 2x=\cos^2x-\sin^2x$ and proceed as in (b) above. (d) Use $2\sin(x/2)\cos(x/2)=\sin x$ and $\cos^2(x/2)+\sin^2(x/2)=1$ to find $\sin(x/2),$ together with the caveat that $\sin(x/2)$ must be positive since $x/2$ must fall within the second quadrant since if $180°<x<270°,$ then $90°<x/2<135°<180°.$
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