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Math Help - Trigonometric Equation involving Tangent and Arctangent

  1. #1
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    Trigonometric Equation involving Tangent and Arctangent

    Dear all,

    For the following equation which I have to solve for t, I am getting an expression that differs from the given answer and I do not understand why.

    I have made two attempts, both posted below. Thank you very much for your help.

    ---

    Show: \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2} \Rightarrow t = \frac{1}{\sqrt{5}}\left(\pi - arctan\frac{2\sqrt{5}}{\alpha + 2}\right), where  \alpha \geq 0

    ---

    Solution 1: \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2}

     \Rightarrow t\sqrt{5} = \arctan\frac{-2\sqrt{5}}{\alpha + 2}

     \Rightarrow t = -\frac{1}{\sqrt{5}}\arctan\frac{2\sqrt{5}}{\alpha + 2} ,

    where I have used the facts that \arctan(-x) = -x and took the arctangent of the original equation.
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  2. #2
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    Re: Trigonometric Equation involving Tangent and Arctangent

    Here's one way, though possibly not the best.


    tan(t\sqrt{5}) = \frac{-2\sqrt5}{a+2}

    -tan(t\sqrt{5}) = \frac{2\sqrt5}{a+2}

    tan(\pi-t\sqrt{5}) = \frac{2\sqrt5}{a+2}

    \pi-t\sqrt{5} = arctan(\frac{2\sqrt5}{a+2})

    -t\sqrt{5} = -\pi+arctan(\frac{2\sqrt5}{a+2})

    t = \frac{1}{\sqrt{5}}(\pi - arctan(\frac{2\sqrt5}{a+2}))

    Also I think you meant arctan(-x)=-arctan(x) for your fact.

    But you may have forgot about -tan(x) = tan(\pi-x)
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  3. #3
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    Re: Trigonometric Equation involving Tangent and Arctangent

    Quote Originally Posted by scherz0 View Post
    Dear all,

    For the following equation which I have to solve for t, I am getting an expression that differs from the given answer and I do not understand why.

    I have made two attempts, both posted below. Thank you very much for your help.

    ---

    Show: \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2} \Rightarrow t = \frac{1}{\sqrt{5}}\left(\pi - arctan\frac{2\sqrt{5}}{\alpha + 2}\right), where  \alpha \geq 0

    ---

    Solution 1: \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2}

     \Rightarrow t\sqrt{5} = \arctan\frac{-2\sqrt{5}}{\alpha + 2}

     \Rightarrow t = -\frac{1}{\sqrt{5}}\arctan\frac{2\sqrt{5}}{\alpha + 2} ,

    where I have used the facts that \arctan(-x) = -x and took the arctangent of the original equation.
    What they have done is recognised that \displaystyle \frac{-2\sqrt{5}}{\alpha + 2} < 0, so \displaystyle t\sqrt{5} must be in the second or fourth quadrants. Their solution only has the shift to the second quadrant (by subtracting from \displaystyle \pi).
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  4. #4
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    Re: Trigonometric Equation involving Tangent and Arctangent

    Quote Originally Posted by Prove It View Post
    What they have done is recognised that \displaystyle \frac{-2\sqrt{5}}{\alpha + 2} < 0, so \displaystyle t\sqrt{5} must be in the second or fourth quadrants. Their solution only has the shift to the second quadrant (by subtracting from \displaystyle \pi).
    Thank you very much for your replies.

    @Prove It: Does your reply explain what member, Stro, did:

    tan(t\sqrt{5}) = \frac{-2\sqrt5}{a+2}

    \Rightarrow -tan(t\sqrt{5}) = \frac{2\sqrt5}{a+2}

    \Rightarrow tan(\pi-t\sqrt{5}) = \frac{2\sqrt5}{a+2}

    @Stro: I understand perfectly the algebra which you used. However, how did you think of using \tan(\pi - x) = -\tan(x)? Was it based on what member, Prove It, wrote afterwards? I'm trying to grasp the inner logic behind proving this equation.

    By the way, thank you; I did mean arctan(x) = -arctan(x).
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