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Cos^2(x) + sin^2 (x) =1

  1. Nov 25, 2009 #1
    There's no answer for this in the back of the book and I just want to make sure I did it correctly. If anyone could post their answer, I would appreciate it! Thanks.

    Cos^2(x) + sin^2 (x) =1 for every real value of x.(Pythagorean theorem)
    What real values of x will be a solution to cos^n (x)- sin^n (x) =1, for a given positive integer n?
     
  2. jcsd
  3. Nov 25, 2009 #2

    Mark44

    Staff: Mentor

    Re: Question!

    What values did you get?
     
  4. Nov 25, 2009 #3
    Re: Question!

    x=0.64 + pi/4 +/- 2*pi*n, 0.64 + 3pi/4 +/- 2*pi*n
     
  5. Nov 26, 2009 #4

    Mark44

    Staff: Mentor

    Re: Question!

    Can you show me how you got those numbers? BTW, they are not correct.
     
  6. Nov 26, 2009 #5

    Mentallic

    User Avatar
    Homework Helper

    Re: Question!

    What's interesting is that the solutions to x are the same for all odd, and the other set of solutions are equal for all even n.
    Your answers are incorrect. Just test them and you'll see.

    There is one solution for x that works with all positive n, and that is x=0.
     
  7. Nov 26, 2009 #6

    Mark44

    Staff: Mentor

    Re: Question!

    A more accurate way to say this is that there is one set of solutions for all odd integers n, and another set of solutions for all even integers n.
     
  8. Nov 26, 2009 #7
    Re: Question!

    Ok, so if n is even x=0 x=k*pi, and if n is odd x=0 x=3pi/2+2*k*pi???
     
  9. Nov 26, 2009 #8

    Mark44

    Staff: Mentor

    Re: Question!

    For n even, yes. For n odd, what you have is correct, there is another bunch of solutions. What I did was draw graphs of y = cos^n(x) and y = sin^n(x) + 1, and found the points where the two graphs intersect. The graphs of odd powers of sine and cosine look pretty much like the graphs of sine and cosine.
     
  10. Nov 26, 2009 #9

    Mentallic

    User Avatar
    Homework Helper

    Re: Question!

    Yes, thankyou :smile:
    I'm stepping into fields I have yet to learn so my terminology would be questionable at best. I just hope that I can get the point across.
     
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