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Homework Help: (sin x)^n-(cos x)^n =1 for all x

  1. Sep 25, 2005 #1
    Find the value of n such that

    (sin x)^n-(cos x)^n =1 for all x

    I really have no idea how to go about it,or even whether such a number exists.Anybody willing to help?
  2. jcsd
  3. Sep 25, 2005 #2


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    Obvious solutions exist where sin x = 1 and cos x = 0 and where sin x = 0 and cos x = 1. You should be able to find others! :)
  4. Sep 25, 2005 #3


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    If it is supposed to hold for ALL x. It should in particular hold for x=0, right?
  5. Sep 25, 2005 #4
    sin(x)^2 + cos(x)^2=1



    sin(x)^n - cos(x)^n=1

    sin(x)^n=1 + cos(x)^n

    Let n=2l

    sin(x)^2l=1 + cos(x)^2l




    1 + cos(x)^2l=(1-cos(x)^2)^l

    With binominal theorem find a value such as,

    (1-b^2)^l=1 + b^2l
    Last edited: Sep 25, 2005
  6. Sep 25, 2005 #5
    That is impossible. cos(x)<1 and sin(x)<1. A value inferior to one raised to any power is always inferior to 1. I think you mean cos(x)^n + sin(x)^n=1. In that case it's pretty simple; first let's proove that n is even;


    -1^n=1, thus n is even.

    Let n=2l

    Let a right triangle with the angle x have sides a,b and c such as c^2=a^2 + b^2.


    (b^2/c^2)^l + (a^2/c^2)^l=1

    a^2l + b^2l/c^2l=1

    Since a^2 + b^2=c^2

    a^2l + b^2l/(a^2 + b^2)^l=1
    a^2l + b^2l=(a^2 + b^2)^l

    By the binominal theorem we know that the only possible value for l; such as


    is 1.

    Thus l=1

    Since n=2l
    Last edited: Sep 25, 2005
  7. Sep 29, 2005 #6
    How about n<0?
    Surely (cosecx)^n-(secx)^n can be 1 for some n when x is not equal to a(pi/2) where a=0,1,2,... ,not for ALL x.
  8. Sep 30, 2005 #7
    How about for all x where cosecx>secx?
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