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Math Help - How can I prove that tan(angle) = opposite/adjacent

  1. #1
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    How can I prove that tan(angle) = opposite/adjacent

    Hello,

    Given the following naming convention:



    It doesn't seam trivial to me to prove that the length of the segment named "tan \theta " is in fact equal to opposite / adjacent, where opposite and adjacent refer to the sides of the triangle inside the circle.



    Can any help with this?

    Thanks
    JL
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  2. #2
    Member HappyJoe's Avatar
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    What are you allowed to use here? For instance, are you allowed to use that tan is defined as sin/cos, and that sin is opposite/hypotenuse and cos is adjacent/hypotenuse?
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  3. #3
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    The title of the thread might be a little confusing. What I am trying to prove here is that the length of the segment named "tan \theta" is actually the same as the value of the trigonometric function tangent(angle).

    By now I actually think I solved the problem though:

    Given the tangent definition "tan = sin / cos" and known that sin = oppos / hyp and cos = adj / hyp -> tan = opposite / adjacent. Now my real question was "How do we know that the length of the segment is equal to oppos / adj" The answer is straight forward if we give a meaning to each side of the triangle. the adjacent side can be thought as the distance covered along the x axis, the opposite side is a difference of height between the ends of the hyp segment, and hypotenuse is therefore the slope which is deffined as height / distance.

    Finally since the radius of the circle is 1 we can assert that the the length of the segment named "tan \theta" is actually equal to the trigonometric function tangent(\theta )

    Does it make sense?

    Thanks for the reply
    JL
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  4. #4
    Member HappyJoe's Avatar
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    Quote Originally Posted by jlzuri View Post
    Does it make sense?
    Possibly. I can't see the pictures you've posted in the OP though, so not sure. Don't you also have this problem?
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  5. #5
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    Using the unit circle, you explain that the right-angle triangle that is formed by the lengths \displaystyle \sin{\theta}, \cos{\theta} is similar to the right-angle triangle formed by the lengths \displaystyle \tan{\theta}, 1.

    That means \displaystyle k\sin{\theta} = \tan{\theta} and \displaystyle k\cos{\theta} = 1, where \displaystyle k is some constant.

    From the second equation, we can see \displaystyle k = \frac{1}{\cos{\theta}} and so \displaystyle \frac{1}{\cos{\theta}}\sin{\theta} = \tan{\theta}, or \displaystyle \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}.

    Now remembering that \displaystyle \sin{\theta} = \frac{O}{H} and \displaystyle \cos{\theta} = \frac{A}{H}, that means

    \displaystyle \tan{\theta} = \frac{\frac{O}{H}}{\frac{A}{H}} = \frac{O}{A}.

    Q.E.D.
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  6. #6
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    I can't see the pictures but I suspect you have a unit circle (center at (0, 0), radius 1) with a line through the origin at angle \theta and a vertical line tangent to that circle at (1, 0). Yes, the length of the segment from (1, 0) to the original line is tan(\theta).

    The point at which the line crosses the circle has coordinates (cos(\theta), sin(\theta)) (that's one common way of defining sine and cosine). That is, the slope of the line is (sin(\theta)- 0)/(cos(\theta)- 0)= tan(\theta). But then, of course, the slope as calculated by using (1, y) (the point where the vertical line at (1, 0) crosses that line) and (0, 0) is (y- 0)/(1- 0)= y= tan(\theta).
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