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Math Help - Hard Questions on Inverse Trigonometry

  1. #1
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    Unhappy Hard Questions on Inverse Trigonometry

    Could someone please demonstrate to me how to solve these equations:

    a) arccos(x) + arccos(x*sqrt(3)) = pi/2

    b) 2arcsin(x*sqrt(6)) + arcsin(4x) = pi/2</SPAN>
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  2. #2
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    Quote Originally Posted by Sunyata View Post
    Could someone please demonstrate to me how to solve these equations:

    a) arccos(x) + arccos(x*sqrt(3)) = pi/2

    b) 2arcsin(x*sqrt(6)) + arcsin(4x) = pi/2</SPAN>
    see this thread ... solution process is the same.

    http://www.mathhelpforum.com/math-he...ease-help.html
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  3. #3
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    Hello, Sunyata!

    (a)\;\;\arccos(x) + \arccos(x\sqrt{3}) \:=\:\frac{\pi}{2}
    Let: . \begin{array}{ccccccccccc}\alpha \;=\; \arccos(x) &\Longrightarrow & \cos\alpha \;=\; x & \Longrightarrow & \sin\alpha \;=\; \sqrt{1-x^2} \\<br />
\beta \;=\; \arccos(x\sqrt{3}) & \Longrightarrow & \cos\beta \;=\; x\sqrt{3} & \Longrightarrow & \sin\beta \;=\; \sqrt{1-3x^2}<br />
\end{array} .[1]


    The equation becomes: . \alpha + \beta \;=\;\frac{\pi}{2}

    Take the cosine: . \cos(\alpha + \beta) \;=\;\cos\left(\tfrac{\pi}{2}\right)

    . . which becomes: . \cos\alpha\cos\beta - \sin\alpha\sin\beta \;=\;0


    Substitute [1]: . (x)(x\sqrt{3}) - (\sqrt{1-x^2})(\sqrt{1-3x^2}) \:=\:0

    . . . . . . x^2\sqrt{3} - \sqrt{1-4x^2 + 3x^4} \:=\:0 \quad\Rightarrow\quad x^2\sqrt{3} \:=\:\sqrt{1 - 4x^2 + 3x^4}

    Square both sides: . 3x^4 \:=\:1 - 4x^2 + 3x^4 \quad\Rightarrow\quad 4x^2 \:=\:1 \quad\Rightarrow\quad x^2 \:=\:\frac{1}{4}

    . . Therefore: . x \;=\;\pm\frac{1}{2}

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  4. #4
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    x = -\frac{1}{2} is an extraneous solution
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  5. #5
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    Absolutely right, skeeter . . . *blush*

    Thanks for the correction.

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  6. #6
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    Hmmm how exactly do u find the domain for x, skeeter and Soroban?
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  7. #7
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    Quote Originally Posted by xwrathbringerx View Post
    Hmmm how exactly do u find the limit, skeeter and Soroban?
    this problem does not involve a limit ... what "limit" are you talking about?
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  8. #8
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    Oops i mean domain for x.

    Sori
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  9. #9
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    start a new thread.
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  10. #10
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    ARGH I get a massive equation for the 2nd question which I cannot solve...
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  11. #11
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    Ummm for the 2nd question, am i MEANT to get 7744x^10 - 3008x^6 + 176 x^5 + 240x^4 - 8x^2 - 8x + 1 = 0 because I keep on getting it, which I've been trying to solve for ages
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