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Math Help - Inverse Trig Functions

  1. #1
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    Inverse Trig Functions

    I'm having trouble understanding this:
    tan[arc cos (square root of 3 / 2) = square root of 3 / 3
    How does this answer come to be?
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  2. #2
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    This question is saying what is the Tangent of the angle who's cosine is \frac{\sqrt{3}}{2}

    So what is the angle who's cosine equals that. Then take the tangent of your answer....

    But even more cleverly, \tan x=\frac{\sin x}{\cos x}

    When cosine equals \frac{\sqrt{3}}{2}, sin = \frac{1}{2}

    So now compute sin over cos
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  3. #3
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    RE: Inverse Trig Functions

    Let's look at the unit circle: File:Unit circle angles.svg - Wikipedia, the free encyclopedia


    As artvandalay said, the first thing the question is asking for is the angle whose cosine is sqrt(3)/2. Knowing that the unit circle coordinates are (cos x, sin x), we can determine that the angle measure whose cosine is sqrt(3)/2 is pi/3. Now from here, the question is asking for the tangent of this angle, which is the sine value over the cosine value. The sine value here is 1/2, and we know the cosine value is sqrt(3)/2. So:

    tan pi/3 = (1/2)/(sqrt(3)/2)
    The twos cancel out, leaving 1/sqrt(3). When we rationalize the function, we get sqrt(3)/3 (multiply the numerator and denominator by sqrt(3) to remove the radical).

    Hope this helps some.
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