# How to solve for theta for this trig question?

## Homework Statement

cos(theta) = -3/4, pi <= theta <= 2pi

## The Attempt at a Solution

I forgot how to do this. How do I use the special triangles to do this question? Do I need to square the sqrt(3) / 2 one or something? Thanks.

## Answers and Replies

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
Try the Pythagorean theorem.

So I use Pythagorean to get the opposite length? (sqrt(7) / 4) ?

haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
So I use Pythagorean to get the opposite length? (sqrt(7) / 4) ?
It depends what you're trying to find. That will help if you want some other trig function of theta, but not much use if you want theta itself.
If it's a numerical value for theta that you want, at some point you will have to use a calculator or whatever to compute an arccos. But the more interesting aspect of the question is getting an answer in the right quadrant. Do you know the relationships between cos(x), cos(-x), cos(pi+x), etc?

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
Not all trig functions require the use of a calculator to evaluate.

haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
Not all trig functions require the use of a calculator to evaluate.
True, but certainly to find arccos in this case.

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
Not really. If you know basic trig values for 45-45-90 and 30-60-90 triangles, solving this problem should be possible without a calculator.

haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
Not really. If you know basic trig values for 45-45-90 and 30-60-90 triangles, solving this problem should be possible without a calculator.
For |cos(theta)| = 3/4? Are you sure about that? If 3/5 I'd agree.

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
I'm pretty sure that zeion has left the building ...

At any rate, given a value for cos(θ), for π ≤ θ ≤ 2π, one can determine in which quadrant θ lies, and thus determine the values of all the other trig functions at angle θ.

Of course, we're still not sure about just what zeion was supposed to do in regards to this problem.

I'm pretty sure that zeion has left the building ...

At any rate, given a value for cos(θ), for π ≤ θ ≤ 2π, one can determine in which quadrant θ lies, and thus determine the values of all the other trig functions at angle θ.

Of course, we're still not sure about just what zeion was supposed to do in regards to this problem.

I'm supposed to

i) Sketch the angle and state the related acute angle
ii) Determine the exact value of each trigonometric ratio

The previous 3 questions I could do because they were all values I could find on the special triangle so they were easy. But I'm not sure how to relate 3/4 to the special triangles?

HallsofIvy
Science Advisor
Homework Helper
Since $\theta$ lies between $\pi$ and $2\pi$, and $cos(\theta)$ is negative, $\theta$ is in the fourth quadrant. Use a calculator to find $cos^{-1}$ of -4/3 and add $\pi$. Equivalently, find $cos^{1}$ of 3/4 and subtract from $2\pi$.

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
Since $\theta$ lies between $\pi$ and $2\pi$, and $cos(\theta)$ is negative, $\theta$ is in the fourth quadrant. Use a calculator to find $cos^{-1}$ of -4/3 and add $\pi$. Equivalently, find $cos^{1}$ of 3/4 and subtract from $2\pi$.
That should be, θ is in the third quadrant.

cos(θ) is positive for θ in the first & fourth quadrants.

cos(θ) is negative for θ in the second & third quadrants.

Last edited:
haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
It is possible that a numerical answer is not required. An expression involving 'arccos(3/4)' may be acceptable. The trick is to get the expression right. arccos is defined to return a value in [0, pi), but the required answer is in [pi, 2pi].