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Thread: how to calculate angle tan(theta) = -4?

  1. #1
    Senior Member Array SCARYdoor's Avatar
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    Default how to calculate angle tan(theta) = -4?

    I'm given this problem: find all angles theta, 0 </= theta </= 2(pi) which satisfy:

    tan (theta) = -4.

    tan theta = sin/cos = y/x = -4/1 or 4/-1

    so the triangle is in the second and fourth quadrant. adjacent side = 1, opposite = 4, hypotenuse = root 17.

    From here I don't know how to continue. I'm not even sure if it's possible to find this angle without a calculator.

    Anyone remember how this works? Thanks.

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    ..-. ..- -.-. -.- Array LastRailway's Avatar
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    tan theta = -4
    theta = arctan (-4)

    and this should be something along the lines of:
    theta = -4 - (-4^3)/3 + (-4^5)/5 - (-4^7)/7 + (-4^9)/9 ...

    Or, according to Google
    arctan (-4) is equal to -1.32581766
    That is, theta < 0, or the problem is not solved.

    I am probably horribly wrong about this.

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    Quote Originally Posted by LastRailway
    That is, theta < 0, or the problem is not solved.
    No, just add 2PI to that number.

    Furthermore it's supposed to be without calculator.

    Also there are 2 answers.

  4. #4
    ..-. ..- -.-. -.- Array LastRailway's Avatar
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    Quote Originally Posted by walfin View Post
    No, just add 2PI to that number.

    Furthermore it's supposed to be without calculator.

    Also there are 2 answers.
    2 answers? How?
    And how would you solve this without calculator?

  5. #5
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    Quote Originally Posted by LastRailway
    2 answers? How?
    Quote Originally Posted by SCARYdoor
    so the triangle is in the second and fourth quadrant
    Quote Originally Posted by LastRailway
    And how would you solve this without calculator?
    Actually I don't know either.

    EDIT: Well he could always draw the triangle and use a protractor for an estimate.
    Last edited by walfin; 8 Mar 2009 at 04:42 AM.

  6. #6
    Confounded Cogitator Array puzzled-observer's Avatar
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    Well, you could just draw sin and cosin on the same axis and, depending on how accurately you draw it, come up with a good estimate by just picking the points where
    sin = 4cos (and then from those, picking the ones where either sin or cosin is in the negative region (but not both)). It's easier than graphing tangent accurately, anyway.

    Edit: Be sure to use the magnitude of sinx = 4*(magnitude)cosx. So maybe drawing the absolute value sin and cosin graphs would be easier.
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