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

    ---

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

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

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

    where I have used the facts that $\displaystyle \arctan(-x) = -x$ 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.


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

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

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

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

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

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

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

    But you may have forgot about $\displaystyle -tan(x) = tan(\pi-x)$
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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 $\displaystyle t$, 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: $\displaystyle \tan(t\sqrt{5}) = \frac{-2 \times \sqrt{5}}{\alpha + 2} \Rightarrow t = \frac{1}{\sqrt{5}}\left(\pi - arctan\frac{2\sqrt{5}}{\alpha + 2}\right)$, where $\displaystyle \alpha \geq 0 $

    ---

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

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

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

    where I have used the facts that $\displaystyle \arctan(-x) = -x$ and took the arctangent of the original equation.
    What they have done is recognised that $\displaystyle \displaystyle \frac{-2\sqrt{5}}{\alpha + 2} < 0$, so $\displaystyle \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 $\displaystyle \displaystyle \pi$).
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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 $\displaystyle \displaystyle \frac{-2\sqrt{5}}{\alpha + 2} < 0$, so $\displaystyle \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 $\displaystyle \displaystyle \pi$).
    Thank you very much for your replies.

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

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

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

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

    @Stro: I understand perfectly the algebra which you used. However, how did you think of using $\displaystyle \tan(\pi - x) = -\tan(x)$? 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 $\displaystyle arctan(x) = -arctan(x)$.
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