Thread: finding positive angles for inverse functions

Hello again,

The question I have been given is to find two positive angles for 1) arcsin(-.37) 2) arccos(-.37) and 3) arctan(-.37) in the range 0<θ<360.
Mainly the wording of the question is confusing me. I started with (sin^-1(−0.37)) and got -21.7 which I converted to 338° which was one correct answer. I then thought to take that angle and subtract 180° to get the 2nd angle in the II Quadrant, but 158 is not the correct answer. I don't understand what is meant by "2 positive angles between 0 and 360". Any suggestions?
Thanks

if , then ... quad III and IV angles, in other words.

welp 202 was the correct answer, not 100% sure why though. What can I google to find out more about this?

The negative sign tells you in which quadrants the angle lies. For , it's the third and fourth quadrants. (Refer to ACTS or CAST or 'All Silly Tom Cats' or whatever other acronym you use.)
Then ignore the negative sign and use your calculator to see that degrees.
Regard this as a shift or a rotation (clockwise or anticlockwise as appropriate) from the horizontal axis.
For the angle in the third quadrant shift, (or rotate anticlockwise) to get an angle of 180+21.72 = 201.72 degrees.
For the angle in the fourth quadrant the rotation will be clockwise to get you an angle of -21.72 or 360-21.72 = 338.28 degrees.

Always shift or rotate from the horizontal axis.

Thank you for your insight BobP, I now understand the question and all questions like it!