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Sin^2(θ/2)+cos^2(θ/2)=1 ?

  1. Oct 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Does [sin(θ/2)-cos(θ/2)]^2 equal 1 for all values of θ?

    I need to figure this out to solve a physics problem.

    2. Relevant equations

    sin^2(A)+cos^2(A)=1

    3. The attempt at a solution

    [sin(θ/2)-cos(θ/2)]^2
    =sin^2(θ/2)+cos^2(θ/2)
    =1

    But this isn't what I get when I put f(x)=[sin(θ/2)-cos(θ/2)]^2 in my graphing calculator. What's going on???
     
  2. jcsd
  3. Oct 11, 2014 #2

    Mark44

    Staff: Mentor

    The above is wrong. (A - B)2 ##\neq## A2 - B2.
     
  4. Oct 12, 2014 #3

    phinds

    User Avatar
    Gold Member
    2016 Award

    Draw a unit circle with a radius at an arbitrary angle and looks at the equation in terms of what it means in that context. This should show you pretty quickly whether it's true or not.
     
  5. Oct 12, 2014 #4
    Hmm I may be wrong, but I think you meant to type "(A -B)2 ≠ A2+A2" (with a + in the second half of the equation) This is the form I gave it in, unless I'm missing something. Is this not correct? I'm getting this from:

    (A-B)2 = (A-B)(A-B) = A2 + AB - AB + (-B)2 = A2 + B2
     
  6. Oct 12, 2014 #5
    Ok I see what I said previously was incorrect!! Because

    (A-B)2 = (A-B)(A-B) = A2 - 2AB + (-B)2 = A2 - 2AB + B2

    NOT what I had earlier. I guess I just need to re-take basic algebra! Thanks Mark!
     
  7. Oct 12, 2014 #6

    Mark44

    Staff: Mentor

    No, this part is wrong: A2 + AB - AB + (-B)2, namely the +AB term.
     
  8. Oct 12, 2014 #7
    Yes, I understand now :D There should have been a -2AB in the middle there. Thanks for making that clear to me!!!
     
  9. Oct 12, 2014 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is fairly easy to show that [itex]cos^2(\theta/2)- sin^2(\theta/2)= cos(\theta)[/itex], not 1.
     
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