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purplemouse

  • 4 years ago

simplify: tan^3(arcsec (x/2)) x= 2sec t sqrt(x^2 -4) = 2 tan t t = arcsec (x/2)

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  1. anonymous
    • 4 years ago
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    |dw:1329633693726:dw|

  2. anonymous
    • 4 years ago
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    it is all in the triangle

  3. anonymous
    • 4 years ago
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    there is a picture where the arc secant is x/2

  4. anonymous
    • 4 years ago
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    find remaining side by pythagoras, get \[\sqrt{x^2-4}\]

  5. anonymous
    • 4 years ago
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    then find the tangent, then cube the result

  6. dumbcow
    • 4 years ago
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    you could also do a straight substitution tan = sqrt(sec^2 -1) tan = sqrt(x^2/4 -1) tan = sqrt(x^2 -4)/2 tan^3 = (x^2-4)sqrt(x^2-4) / 8

  7. purplemouse
    • 4 years ago
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    so I have (x^2 -4)\[\sqrt{(x ^{2}}-4)\]

  8. purplemouse
    • 4 years ago
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    I get it now

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