lLovePhysics
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Also, I dont' really get inverse trig functions, when you calculate arcsin(2/3) on your calculator, it only gives you the value that is within the "conventional domain" of -pi/2<=x<=pi/2 right?
However, when you solve for the angles of a trig solution like in the problem: 3\sin^{2}\theta+10\sin \theta-8=0, you need to look for the reference angles? That is, the angles outside the conventional domain? (2nd quad, in this case)
Can someone come up with a general rule for this? Why do reference angles qualify as a solution when it is outside of the arcsin domain?
However, when you solve for the angles of a trig solution like in the problem: 3\sin^{2}\theta+10\sin \theta-8=0, you need to look for the reference angles? That is, the angles outside the conventional domain? (2nd quad, in this case)
Can someone come up with a general rule for this? Why do reference angles qualify as a solution when it is outside of the arcsin domain?
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