trig

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Find cos(A+B).

cos A=1/3 and sin B=-1/2, with A in quadrant I and B in quadrant IV.

  • trig -

    cos A = 1 / 3

    sin A = + OR - sqrt ( 1 - cos A ^ 2 )

    sin A = + OR - sqrt ( 1 - ( 1 / 3 ) ^ 2 )

    sin A = + OR - sqrt ( 1 - 1 / 9 )

    sin A = + OR - sqrt ( 9 / 9 - 1 / 9 )

    sin A = + OR - sqrt ( 8 / 9 )

    sin A = + OR - sqrt ( 4 * 2 / 9 )

    sin A = + OR - sqrt ( 4 ) * sqrt ( 2 ) / sqrt ( 9 )

    sin A = + OR - 2 * sqrt ( 2 ) / 3

    sin A = + OR - ( 2 / 3 ) * sqrt ( 2 )

    In quadrant I sine are positive so :

    sin A = ( 2 / 3 ) * sqrt ( 2 )


    sin B = - 1 / 2

    sin B = + OR - sqrt ( 1- cos B ^ 2 )

    cos B = + OR - sqrt ( 1 - ( - 1 / 2 ) ^ 2 )

    cos B = + OR - sqrt ( 1 - 1 / 4 )

    cos B = + OR - sqrt ( 4 / 4 - 1 / 4 )

    cos B = + OR - sqrt ( 3 / 4 )

    cos B = + OR - sqrt ( 3 ) / 2

    In quadrant IV cosine are positive so :

    cos B = sqrt ( 3 ) / 2


    cos ( A + B ) = cos A * cos B - sin A * sin B

    cos ( A + B ) = ( 1 / 3 ) * sqrt ( 3 ) / 2 - ( 2 / 3 ) * sqrt ( 2 ) * ( - 1 / 2 )

    cos ( A + B ) = ( 1 / 6 )sqrt ( 3 ) + ( 2 / 6 ) * sqrt ( 2 )

    cos ( A + B ) = ( 1 / 6 ) [ sqrt ( 3 ) + 2 sqrt ( 2 ) ]

    cos ( A + B ) = [ sqrt ( 3 ) + 2 sqrt ( 2 ) ] / 6

  • trig -

    what's all this work?

    A is in QI, so sinA = √8/3
    B is in QIV so cosB = √3/2

    and then as done in the final paragraph

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