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Thread: 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 $\displaystyle \displaystyle \sin{\theta}, \cos{\theta}$ is similar to the right-angle triangle formed by the lengths $\displaystyle \displaystyle \tan{\theta}, 1$.

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

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

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

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

    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 $\displaystyle \theta$ 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 $\displaystyle tan(\theta)$.

    The point at which the line crosses the circle has coordinates $\displaystyle (cos(\theta), sin(\theta))$ (that's one common way of defining sine and cosine). That is, the slope of the line is $\displaystyle (sin(\theta)- 0)/(cos(\theta)- 0)= tan(\theta)$. 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 $\displaystyle (y- 0)/(1- 0)= y= tan(\theta)$.
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