anonymous
  • anonymous
simplify the expression
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ (1-\cot(x))^{2} }{ \cot(x) }\]
anonymous
  • anonymous
@mathstudent55
anonymous
  • anonymous
@carolinar7

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anonymous
  • anonymous
@pooja195, @Australopithecus
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
it doesnt help me
Australopithecus
  • Australopithecus
have you tried expanding (1-cot(x))^2 ??
Australopithecus
  • Australopithecus
show me the work you have done so far
anonymous
  • anonymous
yeah it wasnt working out for me
anonymous
  • anonymous
i dont know how to put up work
Australopithecus
  • Australopithecus
Expansion rule: (a+b)^2 = (a+b)(a+b) = a*a + a*b + b*a + b*b = a^2 + 2ab + b^2
Australopithecus
  • Australopithecus
Just expand the numerator and I will help you from there
anonymous
  • anonymous
|dw:1438196219013:dw|
Australopithecus
  • Australopithecus
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.
Australopithecus
  • Australopithecus
apply the rule and show me what you get
Australopithecus
  • Australopithecus
oops made a mistake
Australopithecus
  • Australopithecus
you messed up and misplaced a negative in your expansion
welshfella
  • welshfella
= 1 - 2 cotx + cot ^ x ---------------- cot x divide each term by cot x note 1 / cot x = tan x
Australopithecus
  • Australopithecus
remember you have: (1-cot(x))^2 so you need to do the following |dw:1438196628184:dw|
anonymous
  • anonymous
so would it be cot(x)+2/
Australopithecus
  • Australopithecus
Note: - * - = + - * + = - + * - = - + * + = +
welshfella
  • welshfella
( a - b)^2 = a^2 - 2ab + b^2
Australopithecus
  • Australopithecus
so writing it all out you have: 1*1 + 1*(-cos(x)) + 1*(-cos(x)) + (-cos(x))*(-cos(x))
Australopithecus
  • Australopithecus
this isnt something you want to just memorize formulas for you should know the process of solving an expansion
anonymous
  • anonymous
so would it be sin(x)cos(x)-2?
Australopithecus
  • Australopithecus
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: \[\frac{1 - 2\cot(x) + \cot(x)^2}{\cot(x)} = \frac{1}{\cot(x)} - \frac{2\cot(x)}{\cot(x)} + \frac{\cot^2(x)}{\cot(x)}\]
Australopithecus
  • Australopithecus
By rule: \[a^{1} = a\] \[\frac{1}{a} = a^{-1}\] and \[a^{0} = 1\] Thus: \[\frac{a}{a} = a^{1}a^{-1} = a^{1 + (-1)} = a^{0} = 1\]
Australopithecus
  • Australopithecus
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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