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Double angle proof's

  1. Sep 3, 2011 #1

    I'm currently trying to convert polar equations to Cartesian equations and vise-versa. I noticed an equation needed a Trig ID I am not familiar with:

    sin(2theta) = 2sin(theta)*cos(theta)

    cos(2theta) = cos^2(theta) - sin^2(theta)

    My apologies if that looks like an eye soar; I'm new here and don't know how to use those fancy characters. Anyway, I don't recall these ID's. I might have not been paying attention in Trig.

    If anyone can explain to me how they are equal I would greatly appreciate it!
  2. jcsd
  3. Sep 3, 2011 #2
    Use the addition (or if you want subtraction) identity.

    sin( a+b ) = sin(a)cos(b) + cos(a)sin(b)
    cos( a+b ) = cos(a)cos(b) - sin(a)sin(b)

    Setting a for theta and b also for theta. See if you can figure it out...

    sin( theta+theta ) = ??
    cos( theta+theta) = ??
  4. Sep 4, 2011 #3
    If a vector is r*Cos(A) i + r*sin(A) j

    and the other is :

    r*Cos(B) i + r*sin(B) j

    and you dot product them

    and A is bigger than B by A-B difference

    r**2 cos A cos B + r**2 Sin A Sin B = r**2 cos( angle between)

    cos A cos B + sin A sin B = cos(A-B) if B=-B

    it is easy to─▒ show that

    Cos(A+B) = cosA cos B - Sin A sin B

    And where sin**2 + cos**2 = 1

    you can find

    Sin(A+B)=Sin A cosB + cos A sin B
    Last edited: Sep 4, 2011
  5. Sep 4, 2011 #4
    Also Cos(90 - (A+B)) = Sin(A+B)
    cos((90-A ) - B)
    Apply result from proof

    cos (90-A) cos(B)+sin (90-A) sin(B) = Sin(A+B)

    SinA cosB+cosA sin B = Sin(A+B)
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