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anonymous

  • one year ago

simplify the expression

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  1. anonymous
    • one year ago
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    \[\frac{ (1-\cot(x))^{2} }{ \cot(x) }\]

  2. anonymous
    • one year ago
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    @mathstudent55

  3. anonymous
    • one year ago
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    @carolinar7

  4. anonymous
    • one year ago
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    @pooja195, @Australopithecus

  5. Australopithecus
    • one year ago
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    http://www.sosmath.com/trig/Trig5/trig5/trig5.html Here is a list of trig identities, just use them to simplify it. As for (1-cot(x))^2 you can expand it, then cancel out terms

  6. anonymous
    • one year ago
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    it doesnt help me

  7. Australopithecus
    • one year ago
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    have you tried expanding (1-cot(x))^2 ??

  8. Australopithecus
    • one year ago
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    show me the work you have done so far

  9. anonymous
    • one year ago
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    yeah it wasnt working out for me

  10. anonymous
    • one year ago
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    i dont know how to put up work

  11. Australopithecus
    • one year ago
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    Expansion rule: (a+b)^2 = (a+b)(a+b) = a*a + a*b + b*a + b*b = a^2 + 2ab + b^2

  12. Australopithecus
    • one year ago
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    Just expand the numerator and I will help you from there

  13. anonymous
    • one year ago
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    |dw:1438196219013:dw|

  14. Australopithecus
    • one year ago
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    so you have: \[\frac{1^2+\cot(x)*1+\cot^2(x)}{\cot(x)}\] looks good to me simplified you have: \[\frac{1+\cot(x)+\cot^2(x)}{\cot(x)}\] Now next step you need to apply the fraction addition rule: \[\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}\] Note: you can only apply this rule when the denominator is the same.

  15. Australopithecus
    • one year ago
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    apply the rule and show me what you get

  16. Australopithecus
    • one year ago
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    oops made a mistake

  17. Australopithecus
    • one year ago
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    you messed up and misplaced a negative in your expansion

  18. welshfella
    • one year ago
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    = 1 - 2 cotx + cot ^ x ---------------- cot x divide each term by cot x note 1 / cot x = tan x

  19. Australopithecus
    • one year ago
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    remember you have: (1-cot(x))^2 so you need to do the following |dw:1438196628184:dw|

  20. anonymous
    • one year ago
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    so would it be cot(x)+2/

  21. Australopithecus
    • one year ago
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    Note: - * - = + - * + = - + * - = - + * + = +

  22. welshfella
    • one year ago
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    ( a - b)^2 = a^2 - 2ab + b^2

  23. Australopithecus
    • one year ago
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    so writing it all out you have: 1*1 + 1*(-cos(x)) + 1*(-cos(x)) + (-cos(x))*(-cos(x))

  24. Australopithecus
    • one year ago
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    this isnt something you want to just memorize formulas for you should know the process of solving an expansion

  25. anonymous
    • one year ago
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    so would it be sin(x)cos(x)-2?

  26. Australopithecus
    • one year ago
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    So for example: (a + b)(c + d) ^ | first step start with a multiply it by c then multiply it by d you get a*c + a*d Now switch to b (a + b)(c + d) ^ | second step multiply b by c then multiply b by d then you get: a*c + a*d + b*c + b*d as for the next step once you have your problem split it into separate fractions makes it easier to visualize: \[\frac{1 - 2\cot(x) + \cot(x)^2}{\cot(x)} = \frac{1}{\cot(x)} - \frac{2\cot(x)}{\cot(x)} + \frac{\cot^2(x)}{\cot(x)}\]

  27. Australopithecus
    • one year ago
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    By rule: \[a^{1} = a\] \[\frac{1}{a} = a^{-1}\] and \[a^{0} = 1\] Thus: \[\frac{a}{a} = a^{1}a^{-1} = a^{1 + (-1)} = a^{0} = 1\]

  28. Australopithecus
    • one year ago
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    so if you had: \[\frac{\sin(x)}{\sin^2(x)} = \sin^1(x)*\sin^{-2}(x) = \sin^{1-2}(x) = \sin^{-1}(x) = \frac{1}{\sin(x)}\]

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