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Math Help - Proving ... equals to pi/2

  1. #1
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    Proving ... equals to pi/2

    How do I prove that

    sin^{-1}(\frac{x}{3}) - 2tan^{-1}sqrt{\frac{3+x}{3-x}} = -\frac{\pi}{2}

    How do I simplify the tan or sin so they would somehow cancel?
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  2. #2
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    You try to make the RHS equal to zero.
    cos(-pi/2) = 0
    So take the cosines of both sides of your original equation.

    Let A = arcsin(x/3)
    And B = 2arctan[sqrt{(x+3)/(x-3)}]

    So your original equation is in the form
    A -B = -pi/2
    Take the cosines of both sides,
    cos(A-B) = 0
    cosAcosB +sinAsinB = 0 ------------***

    A = arcsin(x/3)
    So,
    sinA = x/3 ----------------(i)
    cosA = sqrt(9 -x^2) /3 ------(ii)

    B = 2arctan[sqrt{(x+3)/(x-3)}]
    So,
    sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
    sinB = sqrt(9-x^2) /18 -------------------------(iii)

    cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
    cosB = -x/18 ----------------------------(iv)

    Hence,
    cosAcosB +sinAsinB = 0 ------------***
    [(sqrt(9 -x^2) /3)*(-x/18)] +[(x/3)*(sqrt(9 -x^2) /18)] =? 0
    [-x*sqrt(9 -x^2) /54] +[x*sqrt(9 -x^2) /54 =? 0
    0 =? 0
    Yes.

    Therefore, your original equation is proven.
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  3. #3
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    cosA = sqrt(9 -x^2) /3 ------(ii)

    B = 2arctan[sqrt{(x+3)/(x-3)}]
    So,
    sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
    sinB = sqrt(9-x^2) /18 -------------------------(iii)

    cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
    cosB = -x/18 ----------------------------(iv)

    I got really lost over here... where do u get cosA from? also I was a bit confused witht he arctan part
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  4. #4
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    Quote Originally Posted by circuscircus View Post
    cosA = sqrt(9 -x^2) /3 ------(ii)

    B = 2arctan[sqrt{(x+3)/(x-3)}]
    So,
    sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
    sinB = sqrt(9-x^2) /18 -------------------------(iii)

    cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
    cosB = -x/18 ----------------------------(iv)

    I got really lost over here... where do u get cosA from? also I was a bit confused witht he arctan part
    Quote Originally Posted by circuscircus View Post
    cosA = sqrt(9 -x^2) /3 ------(ii)

    B = 2arctan[sqrt{(x+3)/(x-3)}]
    So,
    sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
    sinB = sqrt(9-x^2) /18 -------------------------(iii)

    cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
    cosB = -x/18 ----------------------------(iv)

    I got really lost over here... where do u get cosA from? also I was a bit confused witht he arctan part
    I see. Sorry. I thought you could follow those because your questions are all a little bit advanced or complicated.

    -------------------------------

    A = arcsin(x/3)
    So,
    sinA = x/3 ----------------(i)
    cosA = sqrt(9 -x^2) /3 ------(ii)

    Draw the right triangle reference of angle A.
    A = arcsin(x/3)
    So, sinA = x/3
    That means the leg opposite A is x
    The hypotenuse is 3
    So, tThe leg adjacent to A is sqrt(3^2 -x^2) = sqrt(9 -x^2)
    Hence, cosA = (adjacent side)/hypotenuse = sqrt(9 -x^2) /3

    ----------------------------------------------------------------
    The arctangent?

    B = 2arctan[sqrt{(x+3)/(x-3)}]

    , My! I wrote it incorrectly. It should have been
    B = 2arctan[sqrt{(3+x)/(3-x)}]
    as shown in your original equation.

    Zeez, I think I'm getting older. Transferring from scratch paper to computer needs some thinking too. And if you're in a hurry and you're getting older, then your mind cannot catch up easily.

    Okay, let angle C = arctan[sqrt(3+x) /sqrt(3-x)]

    Draw the right triangle reference of angle C.
    C = arctan[sqrt(3+x) /sqrt(3-x)]
    So, tanC = sqrt(3+x) /sqrt(3-x)
    That means the leg opposite C is sqrt(3+x)
    The leg adjacent to C is sqrt(3-x)
    So, the hypotenuse is sqrt[(sqrt(3+x))2 +(sqrt(3-x))^2]
    = sqrt[(3+x) +(3-x)]
    = sqrt(6)

    Hence,
    sinC = sqrt(3+x) /sqrt(6)
    cosC = sqrt(3-x) /sqrt(6)

    So,
    sinB = 2sin(arctan[sqrt{(3+x)/(3-x)}]) * cos(arctan[sqrt{(3+x)/(3-x)}])
    sinB = 2[sqrt(3+x) /sqrt(6)]*[sqrt(3-x) /sqrt(6)]
    sinB = 2[sqrt((3+x)(3-x)) /6]
    sinB = sqrt(9 -x^2) /3 -------------------------(iii)

    cosB = cos^2(arctan[sqrt{(3+x)/(3-x)}]) -sin^2(arctan[sqrt{(3+x)/(3-x)}])
    cosB = [sqrt(3-x) /sqrt(6)]^2 -[sqrt(3+x) /sqrt(6)]^2
    cosB = (3-x)/6 -(3+x)/6
    cosB = -2x/6
    cosB = -x/3----------------------------(iv)

    --------------------------------------------------------------

    Hence,
    cosAcosB +sinAsinB = 0 ------------***
    [(sqrt(9 -x^2) /3)*(-x/3)] +[(x/3)*(sqrt(9 -x^2) /3)] =? 0
    [-x*sqrt(9 -x^2) /9] +[x*sqrt(9 -x^2) /9 =? 0
    0 =? 0
    Yes.

    Therefore, your original equation is proven.

    Whew!
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  5. #5
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    Hello, circuscircus!

    Let me give it a try . . .


    Prove: . \sin^{-1}\!\left(\frac{x}{3}\right) - 2\!\cdot\!\tan^{-1}\!\left(\sqrt{\frac{3+x}{3-x}}\right) \;= \;-\frac{\pi}{2}
    Let: . \alpha \:=\:\sin^{-1}\!\left(\frac{x}{3}\right)\;\text{ and }\;\beta \:=\:\tan^{-1}\!\left(\sqrt{\frac{3+x}{3-x}}\right) \quad\Rightarrow\quad \alpha - 2\beta \:=\:-\frac{\pi}{2}


    We have: . \sin\alpha \:=\:\frac{x}{3} \:=\:\frac{opp}{hyp} . . . and Pythagorus gives us: . adj \:=\:\sqrt{9-x^2}
    Hence: . \cos\alpha \:=\:\frac{\sqrt{9-x^2}}{3}


    We have: . \tan\beta \:=\:\frac{\sqrt{3+x}}{\sqrt{3-x}}\:=\:\frac{opp}{adj} . . . and Pythagorus gives us: . hyp \:=\:\sqrt{6}
    Hence: . \sin\beta \:=\:\frac{\sqrt{3+x}}{\sqrt{6}},\;\;\cos\beta\:=\  :\frac{\sqrt{3-x}}{\sqrt{6}}


    The equation is: . \alpha - 2\beta \:=\:-\frac{\pi}{2}

    Take the sine of the left side: . \sin(\alpha - 2\beta) \;=\;\sin(\alpha)\cos(2\beta) - \sin(2\beta)\cos(\alpha)

    . . = \;\sin(\alpha)\left[1-2\sin^2\!(\beta)\right] - \left[2\sin(\beta)\cos(\beta)\right]\cos(\alpha)

    . . = \;\frac{x}{3}\left(1 - 2\!\cdot\!\frac{3+x}{6}\right) - 2\left(\frac{\sqrt{3+x}}{\sqrt{6}}\right)\left(\fr  ac{\sqrt{3-x}}{\sqrt{6}}\right)\left(\frac{\sqrt{9-x^2}}{3}\right)

    . . = \;\frac{x}{3}\left(-\frac{x}{3}\right) - 2\left(\frac{\sqrt{9-x^2}}{6}\right)\left(\frac{\sqrt{9-x^2}}{3}\right)

    . . = \;-\frac{x^2}{9} - \frac{9-x^2}{9} \;=\; \frac{-9}{9} \;=\;-1


    Since \sin(\alpha - 2\beta) \:=\:-1, then: . \alpha - 2\beta \;=\;-\frac{\pi}{2}

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