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Math Help - Calculating values when combining the tangens and cosinus function

  1. #1
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    Thumbs up Calculating values when combining the tangens and cosinus function

    Hello, I'm currently studying for an exam I will have right before semester 2 starts now in August and I could really use some help

    The question is basically "Calculate the value of tan(cos^-1(0.5))"

    I can't understand why or how I would do this, do I turn cosinus into something using sinus and tangens or where do I begin??

    Thank you!
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    Re: Calculating values when combining the tangens and cosinus function

    Quote Originally Posted by Rask View Post
    Hello, I'm currently studying for an exam I will have right before semester 2 starts now in August and I could really use some help

    The question is basically "Calculate the value of tan(cos^-1(0.5))"
    First, cos^{-1}(0.5). Ask yourself this...The cosine of what angle is 0.5? (Hint: It's one of the "common" angles.)

    -Dan
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    Re: Calculating values when combining the tangens and cosinus function

    Alright, so cos(0.5) = 60deg = 1pi/3

    tan(60) = sqrt(3) correct?

    and how is this calculated without a calculator, is it possible or do I have to memorize it?
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    Re: Calculating values when combining the tangens and cosinus function

    Quote Originally Posted by Rask View Post
    Alright, so cos(0.5) = 60deg = 1pi/3

    tan(60) = sqrt(3) correct?

    and how is this calculated without a calculator, is it possible or do I have to memorize it?
    It's a good idea to know the sine, cosine, and tangents of 0, pi/6, pi/4, pi/3, pi/2. (0, 30, 45, 60, 90 in degrees.) They come up all the time in Trig.

    One correction: cos(0.5) is not 60. You meant that cos(60) = 0.5 Otherwise, yes, you have it right!

    -Dan
    Last edited by topsquark; August 3rd 2014 at 02:33 PM. Reason: Typo
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    Re: Calculating values when combining the tangens and cosinus function

    Quote Originally Posted by Rask View Post
    Hello, I'm currently studying for an exam I will have right before semester 2 starts now in August and I could really use some help

    The question is basically "Calculate the value of tan(cos^-1(0.5))"

    I can't understand why or how I would do this, do I turn cosinus into something using sinus and tangens or where do I begin??

    Thank you!
    tan(x)= sin(x)/cos(x) so tan(arcos(x))= sin(arcos(x))/cos(arcos(x)). The denominator is just cos(arcos(x))= x. For the numerator, sin^2(x)+ cos^2(x)= 1 so that sin(x)= \pm\sqrt{1- cos^2(x)} and sin(arcos(x))= \pm\sqrt{1- cos^2(arcos(x)}= \pm\sqrt{1- x^2}. The sign is the sign on x itself.

    A less "rigorous" way to do this is to imagine a right triangle with hypotenuse of length 1 and "near side" of length x (so that the cosine of the angle is x/1= x). By the Pythagorean theorem the "opposite side" has length \sqrt{1- x^2}. tangent is "opposite side over near side" so \sqrt{1- x^2}/x. Again, if you are allowing angles outside 0 to [itex]\pi/2[/itex], you will need to check the sign.
    Last edited by HallsofIvy; August 3rd 2014 at 01:46 PM.
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