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Math Help - Cosine A

  1. #1
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    Cosine A

    Our teacher gave the class a challenge problem.

    If z = tan(A/2), show that cos(A) = (1 - z^2)/(1 + z^2)

    I know that replacing z^2 with tan(A/2) is a start but have no idea where to go from there.
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  2. #2
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    Quote Originally Posted by magentarita View Post
    Our teacher gave the class a challenge problem.

    If z = tan(A/2), show that cos(A) = (1 - z^2)/(1 + z^2)

    I know that replacing z^2 with tan(A/2) is a start but have no idea where to go from there.
    \cos A = \cos \left( 2 \left[ \frac{A}{2} \right] \right) = 2 \cos^2 \left( \frac{A}{2} \right) - 1.

    \tan \left( \frac{A}{2} \right) = z \Rightarrow \cos \left( \frac{A}{2} \right) = \frac{1}{\sqrt{z^2 + 1}}.

    Therefore \cos A = \frac{2}{z^2 + 1} - 1 = ....
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  3. #3
    Senior Member nikhil's Avatar
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    Hope you know the identity
    cos2A=(cosA)^2-(sinA)^2 which is equal to
    cos2A=[(cosA)^2-(sinA)^2]/1
    but (cosA)^2+(sinA)^2=1 therfor
    cos2A=[(cosA)^2-(sinA)^2]/[(cosA)^2+(sinA)^2]
    now dividing both numerator and denominator by (cosA)^2
    we get
    cos2A=[(1-(tanA)^2]/[(1+(tanA)^2]
    now put A/2 in place of A and you get
    cosA=[(1-(tanA/2)^2]/[(1+(tanA/2)^2]
    but tanA/2=z therfor

    cosA=[(1-z^2]/[(1+z^2]
    hence proved
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  4. #4
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    Better...

    Quote Originally Posted by nikhil View Post
    Hope you know the identity
    cos2A=(cosA)^2-(sinA)^2 which is equal to
    cos2A=[(cosA)^2-(sinA)^2]/1
    but (cosA)^2+(sinA)^2=1 therfor
    cos2A=[(cosA)^2-(sinA)^2]/[(cosA)^2+(sinA)^2]
    now dividing both numerator and denominator by (cosA)^2
    we get
    cos2A=[(1-(tanA)^2]/[(1+(tanA)^2]
    now put A/2 in place of A and you get
    cosA=[(1-(tanA/2)^2]/[(1+(tanA/2)^2]
    but tanA/2=z therfor

    cosA=[(1-z^2]/[(1+z^2]
    hence proved
    A much better reply and easier to follow.
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