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Proving identities

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data

    prove that cos ((pi/2)-x) = sinx

    2. Relevant equations



    3. The attempt at a solution

    i extended it to: (cos pi/2) (cos -x) + (sin pi/2) (sin -x)
    =1-sinx
     
  2. jcsd
  3. Sep 11, 2008 #2

    Dick

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    i) cos(a-b)=cos(a)cos(b)+sin(a)sin(b). ii) cos(pi/2)=0. Where did that 1 come from?
     
  4. Sep 11, 2008 #3
    i got the 1 from the sin of pi/2.....isnt that 1?
     
  5. Sep 11, 2008 #4
    You cannot expand trig identities like that.

    It's not like [tex]x^2+x=x(x+1)[/tex]

    [tex]\sin{(x+2)}\neq\sin x+\sin2[/tex]

    Have you learned the Sum and Differences formula?

    You can also prove this through triangles.
     
  6. Sep 11, 2008 #5
    yea we have the sum and difference identities
     
  7. Sep 11, 2008 #6

    Dick

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    Ok, so 1-sinx actually means 1*(-sin(x))?? That isn't the clearest way to write it, wouldn't you agree?? You still have a sign error.
     
  8. Sep 11, 2008 #7
    yea ur right...i forgot the brackets...but it still come sout at -sin(x)......
     
  9. Sep 11, 2008 #8
    oh wait....do i need to include the - on the x?
     
  10. Sep 11, 2008 #9
    cause the subtraction formula is cos ( x - y), and the part of the formula im using is sinxsiny, so do i just need the y number?
     
  11. Sep 11, 2008 #10

    Dick

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    Look at the second post. You have a sign error in cosine sum rule.
     
  12. Sep 11, 2008 #11
    Are you familiar with even and odd functions? It's the same with trig functions.

    even: f(x)=f(-x)

    odd: f(-x)=-f(x)
     
  13. Sep 11, 2008 #12

    Dick

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    Yes. You just need the 'y number'.
     
  14. Sep 11, 2008 #13
    ah ****in eh....thanks guys
     
  15. Sep 11, 2008 #14
    and rocomath, i tried it with the odd even funtions and i got :

    -sinx, because its an odd number infront of the pi/2, and feta=-x....am i missing something?
     
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