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Math Help - Need help with prooving

  1. #1
    Newbie
    Joined
    Apr 2006
    Posts
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    Need help with prooving

    Im not a very good mathematician, and im only in grade 10..and need some help

    Question is: Show
    sin(180-sinB)
    ____________ = -tanB
    cos(180-B)

    Sorry if this question is really easy, i must really suck.
    Ive tried doing it:

    sinB
    ____ = tan B
    cosB

    .: sin B- sinx
    ___________ = -tanx
    cos B- cosx

    Is this method right, and in an exam will i get full marks for it?

    Another question:
    Simplify cos(90-b)
    _________
    sin(180-b)

    I havent the slightest idea what to do.
    I understand that cosx over sinx= 1/2 x tanx..but the values of cos and sin are different in this question... so is it implying that the answer is:

    cos90- cos b
    _____________ = 1/4 x tan (since cos over sin is 1/2tan, this is half)
    sin 180- sin b

    And i need to solve for b....

    Thanks in advance.
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  2. #2
    Grand Panjandrum
    Joined
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    Quote Originally Posted by xanith_47
    Im not a very good mathematician, and im only in grade 10..and need some help

    Question is: Show
    sin(180-B)
    ____________ = -tanB
    cos(180-B)
    Use the difference formulas for \cos and \sin:

    <br />
\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)<br />

    and:

    <br />
\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)<br />

    So if A=180^\circ:

    <br />
\sin(180-B)=\sin(180)\cos(B)-\cos(180)\sin(B) =\sin(B)<br />

    <br />
\cos(180-B)=\cos(180)\cos(B)+\sin(180)\sin(B) =-\cos(B)

    Hence:

    <br />
\frac{\sin(180-B)}{\cos(180-B)}=\frac{\sin(B)}{-\cos(B)}=-tan(B)<br />

    RonL
    Last edited by CaptainBlack; May 12th 2006 at 05:12 AM.
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