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Math Help - Need help with trigonometri

  1. #1
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    Need help with trigonometri

    solve equation: 2arcsinx = arccos(2x+x^2)


    determine on easyest way posible the number:
    2arctan7 + arctan(31/17).
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  2. #2
    MHF Contributor
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    Hi

    2 \arcsin x = \arccos(2x+x^2)

    \cos(2 \arcsin x) = \cos(\arccos(2x+x^2))

    1-2\sin^2(\arcsin x) = \cos(\arccos(2x+x^2))

    1-2x^2 = 2x+x^2

    x=-1 or x=\frac{1}{3}

    You need to verify if those solutions are real solutions
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  3. #3
    Super Member

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    Hello, johnnyboiy!

    Hmmm, there must be an easier way . . .


    Determine in the easiest way possible the number:
    . . \theta\;=\;2\arctan7 + \arctan\left(\tfrac{31}{17}\right)
    \text{We are dealing with two angles: }\;\theta\;=\;2\underbrace{\arctan7}_{\alpha} + \underbrace{\arctan\left(\tfrac{31}{17}\right)}_{\  beta}

    And we will determine: . \tan\theta \;=\;\tan(2\alpha + \beta)\;=\;\frac{\tan(2\alpha) + \tan(\beta)}{1 - \tan(2\alpha)\tan(\beta)} .[1]



    We have: . \alpha \,=\,\arctan7 \quad\Rightarrow\quad \tan\alpha \,=\,7

    Then: . \tan(2\alpha) \:=\:\frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \:=\:\frac{2(7)}{1 - 7^2} \quad\Rightarrow\quad\tan(2\alpha) \:=\:-\frac{7}{24} .[2]


    Since \beta \,=\,\arctan\left(\frac{31}{17}\right), we have: . \tan(\beta) \,=\,\frac{31}{17} .[3]


    Substitute [2] and [3] into [1]: . \tan\theta \;=\;\frac{-\dfrac{7}{24} + \dfrac{31}{17}}{1 - \left(-\dfrac{7}{24}\right)\left(\dfrac{31}{17}\right)} \;=\;\frac{\left(\dfrac{625}{408}\right)}{\left(\d  frac{625}{408}\right)} \;=\;1

    Therefore:. \tan\theta \:=\:1 \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{4}

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