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Math Help - Tough question

  1. #1
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    Tough question

    I'm not strong with this, i need some help:

    2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180
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  2. #2
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    Quote Originally Posted by OmegaCenturion View Post
    I'm not strong with this, i need some help:

    2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180
    Your clue lies in "b+c=180 degrees", as now c=180-b.

    sin(angle) gives the vertical co-ordinate in a circle of radius 1, centre (0,0).

    Hence sin(b)=sin(180-b).

    Therefore your equation becomes

    2\left(1-sin^2(b)\right)=cos^2b+cos^2c

    Using similar logic

    cos(c)=-cos(b)

    allows you to write the entire equation in terms of any one of the four of
    sinc, sinb, cosc, cosb

    with cos^2b+sin^2b=1 etc
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by OmegaCenturion View Post
    I'm not strong with this, i need some help:

    2(1-sin(b)sin(c)) = cos^2(b) + cos^(c) when b+c = 180
    2(1-sinb. sinc) = cos^{2}b + cos^{2}c

    write this as:

     2 - 2.sinb.sinc =  (1-sin^{2}b) + (1-sin^{2}c)

    or,

    2 - 2.sinb.sinc = 2 - sin^{2}b - sin^{2}c

    sin^{2}b - 2.sinb.sinc + sin^{2}c = 0

    (sinb -sinc)^2 = 0

    Note that:

    b+c = 180 \rightarrow c=180-b

    \therefore (sinb -sinc) = (sinb - sin(\pi-b)) = (sinb-sinb) = 0

    You should know that sin(180-b) = sin(\pi-b) = sinb
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  4. #4
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    thanks guys!
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