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Trigonometry question find range

  1. Sep 9, 2008 #1
    guys , following is the problem

    (sinx)^4 + (cosx)^4


    (sinx)^6 + (cosx)^6

    they are asking for the range for each of them

    i really have know clue how to solve this...any sugeestion please?....thanks for reading this!!
  2. jcsd
  3. Sep 9, 2008 #2
    Both of them can be solved in the same way I think. Here is a hint on how to transform the first one.

    [tex]a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2[/tex]
  4. Sep 9, 2008 #3
    so the other one will be

    (a+b)^6 =(a^2 +b^2)^3- 3a^2b - 3ab^2


    i tried to solve and factorize the first one but can`t

    it becomes 2(sin^2(x))^2 - sin^2(x) + 1 ??
    Last edited: Sep 9, 2008
  5. Sep 9, 2008 #4
    Use the property [tex]sin(2x) = 2sin(x)cos(x)[/tex]
  6. Sep 9, 2008 #5
    thanks praharmita... but there is not sin 2x ?

    it`s ((sinx)^2)^2....
  7. Sep 9, 2008 #6
    you dont have to write everything in terms of sin(x) as you did. Use the original form as given by snipez90, where there is a [tex]2a^2b^2[/tex] term, and then use the identity of sin(2x)
  8. Sep 10, 2008 #7
    ooo i see...thanks again....

    is the ans 90<x<270 for the first one ?
  9. Sep 10, 2008 #8
    I assume you mean degrees, in which case you are thinking about the domain but we want to focus on the range (the outputs of our function).

    We have

    [tex]sin^4(x) + cos^4(x) = (sin^2(x) + cos^2(x))^2 - 2sin^2(x)cos^2(x) = 1 - [2sin(x)cos(x)]sin(x)cos(x) = 1 - sin(2x)(\frac{1}{2}sin(2x)) = 1 - \frac{1}{2}sin^2(2x)[/tex]

    Now using what you know about the range of sin and the nature of squared quantities, determine the range of the final expression.
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