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Math Help - evaluating trig inverse functions

  1. #1
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    evaluating trig inverse functions

    arccos( negative square root of 3 / 2)

    i dont understand how you use the unit circle to figure out these kind of problems
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  2. #2
    Senior Member TriKri's Avatar
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    Do you mean arccos\left(-\sqrt{\frac{3}{2}}\right)? In that case it won't end up in a real number. -\sqrt{\frac{3}{2}} is less than -1, hence you can't apply the real arccos function. The point would fall outside the circle, and I know, that is impossible. You will have to use complex numbers.
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  3. #3
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    no, it's square root of 3 over 2.

    and the answer is 5pi/6, why? i tried the unit circle thing and i dont get it. do you follow the x's ---> (x,y) coordinates of the unit circle? i used the ASTC, all students take calculus and i know it must be in either quadrant 2 or 3 because they are the only quads in which cos is negative. so why did they pick quad 2?
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  4. #4
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  5. #5
    Senior Member TriKri's Avatar
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    Oh, you mean arccos\left(-\frac{\sqrt{3}}{2}\right). I think it is just a question of recognizing standard angles.

    Say that arccos\left(-\frac{\sqrt{3}}{2}\right)=\alpha, it means that cos(\alpha)=-\frac{\sqrt{3}}{2}. And since you know that cos^2(\alpha) + sin^2(\alpha) = 1, sin^2(\alpha) = 1 - \frac{3}{4} = \frac{1}{4}. Now, you know that sin(\alpha) = \pm 0.5, maybe that angle is easier to recognize. \alpha is in the interval [0,\ \pi], so sin(\alpha) \geq 0, and then \alpha is either \frac{\pi}{6} or \frac{5\pi}{6}. And since cos(\alpha) = -\frac{\sqrt{3}}{2} < 0, the angle has to be \frac{5\pi}{6}.

    The case when sin(\alpha) = 0.5 is extra special; if you draw a line from (0,\ 0) to (0,\ 1), a line from (0,\ 0) to (cos(\alpha),\ sin(\alpha)), and a line from that point to (1,\ 1), you'll form a triangle. Since the lines beginning in (0, 0) are both radiuses, the have equal length. And since sin(\alpha) = 0.5 which is right between 0 and 1, then the distance from (cos(\alpha),\ sin(\alpha)) must be equal to (0,\ 0) and to (0,\ 1), so all three lines must be of equal length and you have drawn a equilateral, which is known for having all angles being \frac{\pi}{3} radians. Hence sin(\alpha) must be either \frac{\pi}{2}+\frac{\pi}{3} or \frac{\pi}{2}-\frac{\pi}{3}.
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