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Trignometric identiy (LOADS o HELP ;D )

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data

    sin(pi +A) sec(pi-A)
    --------- + ---------- = 0
    cos (2pi + A) csc(2pi-A)

    2. Relevant equations
    cos(x+y)= cosxcosy-sinxsiny
    cos(x-y)= cosxcosy+sinxsiny

    3. The attempt at a solution
    sin pi cosA + cosA sin pi || 1/cos pi cos A + sin pi sin A
    ------------------------ + --------------------------
    cos 2pi cosA + sin 2pi sinA || 1/sin pi cos A - cos pi sin A

    the || is a seperator of the fractions

    if it helps, i got this one to equal zero :)
  2. jcsd
  3. Dec 8, 2009 #2


    Staff: Mentor

    First off, use the identities that involve the sum of pi and an angle. E.g., sin(pi + A) = -sin(A) and cos(pi + A) = -cos(A). Also, cos(2pi + A) = cos(A) and so on.

    Since all of the trig functions are periodic with period 2pi, f(2pi + A) = f(A), where f represents sine, cosine, tangent (yes I know that tangent's period is pi), and all the rest.

    In the work you did, you have sin(pi) and cos(2pi) and all. Simplify those to 0 and 1, respectively.
  4. Dec 10, 2009 #3
    Is this your equation?

    [itex]\frac {\sin (\pi+A)}{\cos(2\pi+A} \ + \ \frac{\sec(\pi-A)}{\csc(2\pi+A)} \ = \ 0[/itex]
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