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Thread: Find the exact value of tan -105 degrees

  1. #1
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    Find the exact value of tan -105 degrees

    tan -105 degrees

    Edit: Never mind I got it, the answer is 2+2radical3, correct?

    Edit 2: Forgot to cancel the 2 before the radical, so it's 2+radical 3...
    Last edited by toeknee; Feb 28th 2011 at 08:15 PM.
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  2. #2
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    $\displaystyle \displaystyle \tan 105 = \tan (60+45)$

    Now apply

    $\displaystyle \displaystyle \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$
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  3. #3
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    An angle of $\displaystyle \displaystyle 105^{\circ}$ in the clockwise direction is the same as an angle of $\displaystyle \displaystyle 255^{\circ}$ in the anticlockwise direction.

    So $\displaystyle \displaystyle \tan{\left(-105^{\circ}\right)} = \tan{\left(255^{\circ}\right)}$

    $\displaystyle \displaystyle = \tan{\left(180^{\circ} + 75^{\circ}\right)}$

    $\displaystyle \displaystyle = \tan{\left(75^{\circ}\right)}$ since the tangent values are the same in the first and third quadrants...

    $\displaystyle \displaystyle = \tan{\left(30^{\circ} + 45^{\circ}\right)}$.


    Now make use of the identity $\displaystyle \displaystyle \tan{\left(\alpha + \beta\right)} \equiv \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}}$.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    $\displaystyle \displaystyle \tan 105 = \tan (60+45)$

    Now apply

    $\displaystyle \displaystyle \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$
    Yes it is, but the OP is evaluating $\displaystyle \displaystyle \tan{\left(-105^{\circ}\right)}$...
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  5. #5
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    Good pick up Hayden, I didn't see that little fella in there, my eyes must be failing me...

    I hate being old.
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    Quote Originally Posted by Prove It View Post
    Yes it is, but the OP is evaluating $\displaystyle \displaystyle \tan{\left(-105^{\circ}\right)}$...
    But $\displaystyle \tan\left(-x\right) = -\tan\left(x\right)$, so $\displaystyle \tan\left(-105^{\circ}\right) = -\tan\left(105^{\circ}\right) = -\cdots $ (Pickslide's post).
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  7. #7
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    Quote Originally Posted by TheCoffeeMachine View Post
    But $\displaystyle \tan\left(-x\right) = -\tan\left(x\right)$, so $\displaystyle \tan\left(-105^{\circ}\right) = -\tan\left(105^{\circ}\right) = -\cdots $ (Pickslide's post).
    I'll pretend I meant that...
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