Thread: Finding two angles between 0 and 360 that have the same trig function values

Hi guys,

I'm new here, hoping to get some help figuring out how to solve these problems.

I am given a trig function and its value, then I am supposed to locate two angles based off that value. I am not really sure where to begin with it.

=

thanks for help all

Originally Posted by mathgex

Hi guys,

I'm new here, hoping to get some help figuring out how to solve these problems.

I am given a trig function and its value, then I am supposed to locate two angles based off that value. I am not really sure where to begin with it.

cos theta = -1/2

thanks for help all

First we need to we need to note that cosine is negative in quadrants II and III.

We can then draw a reference triangle in each of those quadrants using the fact that cosine is the adjacent side over the hypotenuse. This will construct two of our special Triangles. Now that we have the reference angle, we can find the other two angles. Draw a diagram and post back if you get stuck.

Originally Posted by TheEmptySet

First we need to we need to note that cosine is negative in quadrants III and IV.

We can then draw a reference triangle in each of those quadrants using the fact that cosine is the adjacent side over the hypotenuse. This will construct two of our special Triangles. Now that we have the reference angle, we can find the other two angles. Draw a diagram and post back if you get stuck.

I made a drawing here, not sure I did it correctly tho

Originally Posted by mathgex

I made a drawing here, not sure I did it correctly tho

review how to sketch a reference triangle ...

ok i have a drawing, now what do I do though?

Originally Posted by mathgex

ok i have a drawing, now what do I do though?

re-read EmptySet's response.

I'm not sure what he is saying.. if I could just see a similar problem solved once step by step I can build an association for everything.

Originally Posted by mathgex

I'm not sure what he is saying.. if I could just see a similar problem solved once step by step I can build an association for everything.

note that the reference triangles are 30-60-90 degree right triangles.

Originally Posted by skeeter

note that the reference triangles are 30-60-90 degree right triangles.

Ok, so...

I took cos^-1(-.5) and got 120 degrees.

So I there is the first angle right?

Some how before I had found that cos(240) = -.5

So would that be the second angle? I am not sure why though because the only relationship I found between 240 and 120 is that 120 is 1/2 of 240.

Does that mean that I just find the first angle each time and then double it or cut it in half to get the second? I feel like it should be harder than that.

Originally Posted by skeeter

Idk I am not getting it.


EDiT: maybe I was confusing myself ... is that basically the angles are reflected across the x-axis or y-axis depending on which ever trig function is in question?

cos function would reflection across the x-axis
sin across the y-axis
and tan across both x and y axis

right?

The diagram represents the special right triangle (30-60-90 and 45-45-90) angle values on the unit circle in both degrees and radians.

Each point on the circle has coordinates as shown in the diagram, and yes, symmetry across the x and y axes is prevalent.

I recommend you do further research on the unit circle in your text or on the web. This is basic trig information your instructor should have already covered.

Originally Posted by skeeter

The diagram represents the special right triangle (30-60-90 and 45-45-90) angle values on the unit circle in both degrees and radians.

Each point on the circle has coordinates as shown in the diagram, and yes, symmetry across the x and y axes is prevalent.

I recommend you do further research on the unit circle in your text or on the web. This is basic trig information your instructor should have already covered.

we have covered the unit circle I was just having trouble understanding how all of this goes together. That or I just didn't understand what was being asked of me to do.

Thanks for all your help guys I am confident I understand this bit of the material now.
I maybe back later on for help though