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Math Help - Deriving identities

  1. #1
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    Deriving identities

    Okay these are giving me such a problem to figure out. I have to use these 4 identities to prove others.
    A. sin(-x) = - sinx
    B. cos(-x) = -cosx
    C. cos(x+y) = cosxcosy-sinxsiny
    D. sin(x+y) = sinxcosy + cosxsiny

    these are what i have to derive.

    cos2x=1-2sin^2x

    Abslout value of cos(x/2) = Squareroot of ((1+cosx)/2)

    Abslout value of sin(x/2) = Squareroot of ((1-cosx)/2)

    Any help is much apprecieted. Thanks alot!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ocmisssunshine View Post
    Okay these are giving me such a problem to figure out. I have to use these 4 identities to prove others.
    A. sin(-x) = - sinx
    B. cos(-x) = -cosx
    C. cos(x+y) = cosxcosy-sinxsiny
    D. sin(x+y) = sinxcosy + cosxsiny

    these are what i have to derive.

    cos2x=1-2sin^2x
    Use C with y=x:

    <br />
\cos(x+y)=\cos(2x) = \cos^2(x)-\sin^2(x) <br />

    Now use \cos^2(x)=1-\sin^2(x) to get:

    <br />
\cos(2x) = \cos^2(x)-\sin^2(x)=1-2\sin^2(x) <br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ocmisssunshine View Post
    Okay these are giving me such a problem to figure out. I have to use these 4 identities to prove others.
    A. sin(-x) = - sinx
    B. cos(-x) = -cosx
    C. cos(x+y) = cosxcosy-sinxsiny
    D. sin(x+y) = sinxcosy + cosxsiny

    these are what i have to derive.

    cos2x=1-2sin^2x

    Abslout value of cos(x/2) = Squareroot of ((1+cosx)/2)
    The partner of the previous identity is:

    <br />
\cos(2x)=1+2\cos^2(x)<br />

    put u=2x, so:

    <br />
\cos(u)=1+2\cos^2(u/2)<br />

    Now rearrange:

    <br />
|\cos(u/2)| = \sqrt{(1+\cos(u))/2}<br />

    Now replace u with x:

    <br />
|\cos(x/2)| = \sqrt{(1+\cos(x))/2}<br />


    ]
    Abslout value of sin(x/2) = Squareroot of ((1-cosx)/2)

    Any help is much apprecieted. Thanks alot!
    Do this like the one above but using the result of the first part

    RonL
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