Inverse Trig and Other Quadrants
Date: 11/05/2011 at 19:06:43 From: Steven Subject: How do I find the solutions to an inverse trig function Hi, How do you find both solutions for the inverse of a trigonometric function? For example, say I have sin^-1 (1/2). There are two general solutions: 30 degrees and 150 degrees. But when I evaluate that expression on the calculator, I get only one of the answers. I know that the inverse of a trig function has infinitely many solutions, because you can add any integer of pi or 360 to get the next answer. I just can't figure out how to find both of the angles that are between 0 and 360 or 0 and 2pi. Is there a specific formula to find both angles that are between those values?
Date: 11/05/2011 at 23:18:34 From: Doctor Peterson Subject: Re: How do I find the solutions to an inverse trig function Hi, Steven. Here's the basic idea, boiled down to a standard procedure. First, look at the sign of the given value and determine which quadrants the two solutions will lie in. In your example, the sine of the angle is positive, so the angle will be in either the first or second quadrant. I use these pictures: sin cos tan csc sec cot + | + - | + - | + --+-- --+-- --+-- - | - - | + + | - Second, ignore the sign of the given value and use the inverse function to find the reference angle, which will always be in the first quadrant. Call the reference angle R. Now put these together. For an angle in the ... ... first quadrant, the solution is the reference angle R, itself. ... second quadrant, the solution is 180 - R. ... third quadrant, the solution is 180 + R. ... fourth quadrant, the solution is 360 - R. I use these pictures: Q I Q II Q III Q IV | / \ | | | |/R R\| | | 180----+---- 180----+---- 180----+---- ----+---360 | | R/| |\R | | / | | \ So for your example, the reference angle is 30 degrees. One solution (in the first quadrant, the reference angle) is 30 degrees, and the other (in the second quadrant) is 180 - 30 = 150 degrees. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 11/06/2011 at 14:58:12 From: Steven Subject: Thank you (How do I find the solutions to an inverse trig function) Thank you! The pictures really helped.
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