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Math Help - trig inverse...

  1. #1
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    trig inverse...

    i have two pre calculus problems.please help!
    problem 1: if  \cos ^{-1} x + \cos^{-1}y + \cos^{-1}z=\pi then prove that  x^{2}+y^{2}+z^{2}+2xyz=1

    problem 2:evaluate  \lim_{n\to 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3x-2}}
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  2. #2
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    Quote Originally Posted by earthboy View Post
    i have two pre calculus problems.please help!
    What are the problems?
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  3. #3
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    what i have done:

    problem 1:
    i took  \cos ^{-1} x , \cos^{-1}y , \cos^{-1}z as A,B and C respectively.now  A+B+C=\pi now we have to show  \cos ^{2} A + \cos^{2}B + \cos^{2}C + 2 \cos A\cos B\cos C = 1  but i keep getting stuck after thatplease help!
    problem 2: is the answer 0? i factored the expression into  \frac{(2-x)(\sqrt{x+2}+\sqrt{3x-2})}{2} and substituted 2 in it.

    thanks in advance!
    Last edited by earthboy; January 9th 2011 at 07:02 PM.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    What are the problems?
    sorry i pressed the submit button by mistake...edited
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  5. #5
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    is the answer to the limit problem 0 ?
    please reply fast.
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  6. #6
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    Hello, earthboy!

    \displaystyle\text{2. Evaluate: }\;\lim_{x\to 2} \frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3x-2}}

    Multiply numerator and denominator by the conjugate:

    \displaystyle \frac{x^2-4}{\sqrt{x+2} - \sqrt{3x-2}} \cdot\frac{\sqrt{x+2} + \sqrt{3x-2}}{\sqrt{x+2} + \sqrt{3x-2}} \;=\;\frac{(x^2-4)(\sqrt{x+2} + \sqrt{3x-2})}{(x+2) - ( 3x-2)}


    . . . . \displaystyle =\;\frac{(x^2-4)(\sqrt{x+2} + \sqrt{3x-2})} {-2x + 4} \;=\; \frac{(x-2)(x+2)(\sqrt{x+2} + \sqrt{3x-2})}{-2(x-2)}


    . . . . \displaystyle =\;\frac{(x+2)(\sqrt{x+2} + \sqrt{3x-2})}{-2}



    \displaystyle \text{Therefore: }\;\lim_{x\to2}\frac{(x+2)(\sqrt{x+2} + \sqrt{3x-2})}{-2} \;=\;\frac{4(\sqrt{4} + \sqrt{4})}{-2}

    . . . . . . . . . . \displaystyle =\;\frac{4(2+2)}{-2} \;=\;\frac{16}{-2} \;=\;-8

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  7. #7
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    Quote Originally Posted by earthboy View Post
    problem 1:
    i took  \cos ^{-1} x , \cos^{-1}y , \cos^{-1}z as A,B and C respectively.now  A+B+C=\pi now we have to show  \cos ^{2} A + \cos^{2}B + \cos^{2}C + 2 \cos A\cos B\cos C = 1  but i keep getting stuck after thatplease help!

     \cos ^{2} A + \cos^{2}B + \cos^{2}C + 2 \cos A\cos B\cos C
     = \cos ^{2} A + \cos^{2}B + \cos^{2}(\pi - A - B) + 2 \cos A\cos B\cos(\pi - A - B)
     = \cos ^{2} A + \cos^{2}B + \cos^{2}(A + B) - 2 \cos A\cos B\cos(A + B)
     = \cos ^{2} A + \cos^{2}B + (\cos A \cos B - \sin A \sin B)^2 - 2 \cos A\cos B(\cos A \cos B - \sin A \sin B)
     = \cos ^{2} A + \cos^{2}B + \cos^2 A \cos^2 B - 2\cos A \cos B \sin A \sin B + \sin^2 A \sin^2 B - 2 \cos^2 A\cos^2 B + 2\cos A \cos B \sin A \sin B
     = \cos ^{2} A + \cos^{2}B + \cos^2 A \cos^2 B + \sin^2 A \sin^2 B - 2 \cos^2 A\cos^2 B
    ...

    Now use \sin^2 x  = 1 - \cos^2 x
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