Finding an inverse trig value, like trig-1(A):
Remember: In this type of problem, A is something associated with the terminal point, and we're trying to find the distance around the unit circle for this terminal point
1. Based upon the plus/minus sign of A, find the two quadrants we must be in for our trig function to have that plus/minus sign.
2. Based upon the principal values for the trig function we are using, find the two quadrants we must be in.
3. Identify the correct quadrant we must be in, based on the overlap of (1) & (2).
4. Translate |A| into a terminal point. (i.e. ignore any minus sign on A).
5. Find the 1st quadrant t-value for the terminal point in (4).
6. Adjust the t-value to the correct quadrant, based on (3).

Example: Find sin-1(1/2):
1. Since 1/2 is positive, and our trig function is sine, we must be in quadrants I or II, since that is where sine is positive.
2. Since our trig function is sine, we must be in quadrants I or IV, since those are the principal values for sine.
3. To make both (1) & (2) work, we must be in quadrant I.
4. We translate 1/2 into a terminal point. Since our trig function is sine, and sine is the y-value on the unit circle, our terminal point must be (something, 1/2). We could plug y = 1/2 into the equation for the unit circle, or perhaps we just recognize that the x-value must be x = sqrt(3)/2.
5. The first quadrant t-value that gets us to the point (sqrt(3)/2, 1/2) is π/6.
6. We need to adjust π/6 into the correct quadrant, but the correct quadrant is quadrant I this time, which is where π/6 is. Thus our answer is π/6.

Example: Find tan-1(-1):
1. Since -1 is negative, and our trig function is tangent, we must be in quadrants II or IV, since that is where tangent is negative.
2. Since our trig function is tangent, we must be in quadrants I or IV, since those are the principal values for tangent.
3. To make both (1) & (2) work, we must be in quadrant IV.
4. We translate 1 (since that's the absolute value of -1) into a terminal point. Since our trig function is tangent, and tangent is the y/x, it must equal 1 where the y & x values are the same. The only point (in quadrant I) that does that is (sqrt(2)/2, sqrt(2)/2). That's our terminal point.
5. The first quadrant t-value that gets us to the point (sqrt(2)/2, sqrt(2)/2) is π/4.
6. We need to adjust π/4 into the correct quadrant, which is quadrant IV this time. Moving the distance π/4 into the 4th quadrant gives us a t-value of -π/4. That's our answer.

 

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