Question

cos 2θ = cos^2θ − 1/2

Answer 1

\[\cos2 \theta = \cos^{2} \theta - \frac{1}{2}\]
correct?

Answer 2

yep

Answer 3

u can verify formula by putting some arbitrary value

Answer 4

that you know value of

Answer 5

So start off using the double angle formula for cos2(theta)
\[\cos^{2} \theta - \sin^{2} \theta = \cos^{2} \theta - \frac{1}{2}\]
From there you can cancel out cos^2 on each side and say
\[\sin^{2} \theta = \frac{1}{2} \implies \sin \theta = \pm \frac{1}{4}\]

Answer 6

so the final answer is sin(theta) = pm 1/4

Answer 7

?

Answer 8

You still need to actually get theta. We never solved for theta, merely sin(theta). Right now you have
\[\sin \theta = \frac{1}{4}\] and
\[\sin \theta = -\frac{1}{4}\]Any idea how you would solve for theta from here in each of the two equations?

Answer 9

honestly, I don't know what to do from there, all I know the answer is is asking for (theta) = something, something

Answer 10

Right. So when you need to solve for theta, you need to use inverse trig functions. Inverse trig functions basically cancel out theother regular trig functions

\[\sin^{-1} (\sin \theta) = \theta\] the two cancel out and just leave theta. Works for anything really. For example, if this were just
cos(theta) = 1/2, I could get theta by itself by doing inverse cosine of both sides:
\[\cos^{-1} (\cos \theta) = \cos^{-1}(\frac{1}{2}) \implies \theta = \cos^{-1}(\frac{1}{2})\] and you would solve from there.

So same idea with this problem. We want theta by itself without the sin there, so we inverse sin both sides of each equation.

\[\sin^{-1}(\sin \theta) = \sin^{-1}(\frac{1}{4}) \implies \theta = \sin^{-1}(\frac{1}{4})\]and
\[\sin^{-1}(\sin \theta) = \sin^{-1}(-\frac{1}{4}) \implies \theta = \sin^{-1}(-\frac{1}{4})\]

So you basically get two things, inverse sin of -1/4 and inverse sin of positive 1/4. Was there a domain given for this problem? Like between 0 and 2pi?

Answer 11

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)

Answer 12

I put pi/4,3pi/4,5pi/4,7pi/4 and it wasnt correct

Answer 13

Okay, so we just use k then, gotcha. Alright, well what I gave above can be posted as decimal answers, but we would just add 2kpi to them. So inverse sin -1/4 + 2kpi and inverse sin 1/4 + 2kpi. Those give you two of the possible answers. the last ones we need to use reference angles for. So if you get actual decimal approximations of the 2 values, you get about .253 and -.253. But we can find additional angles with reference angles. For the sin function, if you have a point on the unit circle, a point directly across horizontally from it will have the samevalue. So for us, we get the values .253 and -.253

The two points on the right are the first answers we got and the two answers on the left are ones that can be found by drawing horizontally straight across. In order to get those 2 values, we need to recognize that the point for .253 and the point straight across would have the same measure, except where one has a measure of 0 + angle, the other has a measure of pi minus the angle. As for the 2ndpoint, it starts down at a point that would 2pi - angle and straight across from it would be pi + the same angle. So you end up with 4 points, 2 from the equations, two from drawing straight across and relating the angles compared to the x-axis. You get
inverse sin 1/4 + 2kpi, inverse sin -1/4 + 2kpi, pi - (inverse sin 1/4), pi + (inverse sin -1/4)