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Math Help - Exact Values

  1. #1
    Member classicstrings's Avatar
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    Exact Values

    How would I express tan(270-A) in terms of tanA?

    Would I use the double angle formula for tan? But that gives tan270 which is undefined.?
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  2. #2
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    Quote Originally Posted by classicstrings View Post
    How would I express tan(270-A) in terms of tanA?

    Would I use the double angle formula for tan? But that gives tan270 which is undefined.?
    Hello,

    use the property:

    \tan(270^\circ-A)=\frac{\sin(270^\circ-A)}{\cos(270^\circ-a)} =\frac{\sin(270^\circ) \cdot \cos(A)-\cos(270^\circ) \cdot \sin(A)}{\cos(270^\circ) \cdot \cos(A)-\sin(270^\circ) \cdot \sin(A)}

    Now sin(270) = -1 and cos(270) = 0

    Therefore you get:

    \tan(270^\circ-A)=\frac{-\cos(A)}{-\sin(A)}=\frac{1}{\tan(A)}

    EB
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  3. #3
    Member classicstrings's Avatar
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    Thanks earboth!
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  4. #4
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    Hello, classicstrings!

    How would I express \tan(270^o-\theta) in terms of \tan\theta?

    You're right . . . the compound-angle formula gives us \tan270^o
    We must find a way around this dilemma . . .

    Graph the angle . . .
    Code:
                Y
                |   *Q
                |  /
                |θ/
                |/
        - - - - + - - - - X
               /|O
              /θ|
             /  |
            /   |
          P*    |
                Y'

    \angle XOP = 270^o - \theta\quad\Rightarrow\quad \angle POY' = \theta\quad\Rightarrow\quad \angle QOY = \theta

    Hence: . \tan(270^o - \theta) \:=\:\tan\left(90^o - \theta\right) \:=\:\frac{1}{\tan\theta}

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