Can someone teach me how to find reference angles? one of my problems is arc cosine=.4567 What is the reference angle? If you dont like just giving answers away do a sample problem showing me in depth how to do these kind of problems.
First of all, arccos = .4567 is 62.8 degrees (punch it in your calculator to find out).
It's in quadrant one, therefore you don't need a reference triangle because well, I don't know how to explain it. You just don't.
Say you wanted to find arc cos= a value that gets the angle into quadrant 2.
I cannot explain how come, as my knowledge is little. Check youtube or google. The goal is to get the angle into something the 1st quadrant can handle (90 degrees or under). How come? I don't know.
What you do is do 180 degrees-the angle and that's it. (works also for radians as pi= 180 degrees
For the 3rd quadrant,
angle - 180 degrees
4th,
360 - angle
Say from some trig function with a decimal, you get angle 340 degrees.
Now, we know that's in the 4th quadrant as each quadrant (section of the graphing plane) contains the degrees 270 (could be 271) to 360.
So we use, 360 - 340= 20.
Hope that helps some.
A reference angle is the smallest angle measured from either the +x or -x axis.
For example, the reference angle for 170 degrees is 10 degrees. The reference angle for 330 degrees is 30 degrees. (We don't worry about positive or negative angles here.) Simply enough, the reference angle for 60 degrees is 60 degrees.
We have that $\displaystyle acs(0.4567) = 62.8256^o$. This is a reference angle. So the possible angles this could come from are where the cosine of the angle is positive and 62.8256 degrees is the reference angle.
Well, cosine is positive in quadrants I and IV. So this is a reference angle in quadrants I and IV. So the angle will be either 62.8256 degrees or 360 - 62.8256 = 297.174 degrees. Try both of these in your calculator.
-Dan