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Math Help - trig reduction

  1. #1
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    trig reduction

    Sin(-180-x)
    This is my attempt
    First I used sum/difference formulae
    Sin(A-B) =sinAcosB-cosAsinB
    =sin(-180)cos(x)-cos(-180)sin(x)
    =-sin(180)cos(x) - - cos(180)sin(x)
    =0.cos(x) + (-1)]sin(x)
    =-sin(x)
    My another option was to multiply by (-1)
    After multiply I have sin (180+x) which I knw Sin(x) is negative in third quadrant(not sure about this method though)
    According to my textbook answer is Sin (x)
    Where did I lost it??
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  2. #2
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    Re: trig reduction

    Quote Originally Posted by Bonganitedd View Post
    Sin(-180-x)
    The \sin(x) is an odd function.

    This \sin(-180-x)=-\sin(180+x)=-[\sin(x)\cos(180)+\sin(180)\cos(x)]
    Thanks from Bonganitedd and topsquark
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  3. #3
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    Re: trig reduction

    I now see. Thank you.
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  4. #4
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    Re: trig reduction

    Quote Originally Posted by Plato View Post
    The \sin(x) is an odd function.

    This \sin(-180-x)=-\sin(180+x)=-[\sin(x)\cos(180)+\sin(180)\cos(x)]
    Or even \displaystyle \begin{align*} -\sin{ \left( 180^{\circ} + x \right) } = - \left[ -\sin{ (x) } \right] \end{align*} from the symmetry of the unit circle...
    Thanks from topsquark
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