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Math Help - [SOLVED] if cosA = 4/5; sin <0; cosB = 12/13; 0<B>90, determine sin(A-B)

  1. #1
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    [SOLVED] if cosA = 4/5; sin <0; cosB = 12/13; 0<B>90, determine sin(A-B)

    Please help with this one ... thanks so much

    if cosA = 4/5; sin <0; cosB = 12/13; 0<B<90, determine sin(A-B)

    please note that in the heading tere is an error with the B, it says ; 0<B>90 when it should read 0<B<90, like in the above shown way
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  2. #2
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    if cosA = 4/5; sin < 0; cosB = 12/13; 0<B<90, determine sin(A-B)

    This is only barely a calculation problem. You should be sufficiently familiar with the definitions to pull this off.

    You need the Pythagorean Throrem.

    cos(A) = 4/5; sin < 0 ==> sin(A) = -3/5 Note that this is Quadrant IV

    cos(B) = 12/13; 0<B<90 ==> sin(B) = 5/13

    You need the identity:

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

    Let's see what you get.
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  3. #3
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    Hello, Jono Mathematician

    If \cos A = \frac{4}{5},\;\sin A < 0,\quad \cos B = \frac{12}{13},\;0 < B < 90, . determine: \sin(A-B)
    Identity: . \sin(A - B) \;=\;\sin A\cos B - \sin B\cos A\;\;{\color{blue}[1]}


    We are given: . \boxed{\cos A = \frac{4}{5}},\;\sin A < 0

    Cosine is positive in Quadrants 1 and 4; sine is negative in Quadrants 3 and 4.
    . . Hence, A is in Quadrant 4.

    Since \cos A \:=\:\frac{4}{5} \:=\:\frac{adj}{hyp}, we have: . opp \:=\:\pm 3 (use Pythagorus!)

    Since A is in Quadrant 4, opp = -3
    . . Hence: . \boxed{\sin A \:=\:-\frac{3}{5}}


    We are given: . \boxed{\cos B = \frac{12}{13}}\;\text{ and B} is in Quadrant 1.

    Since \cos B \:=\: \frac{12}{13} \:=\: \frac{adj}{hyp}, we have: . opp \:=\:\pm 5

    Since B is in Quadrant 1, opp = 5
    . . Hence: . \boxed{\sin B \:=\:\frac{5}{13}}


    Substitute into the identity [1]:

    . . \sin(A-B) \;=\;\left(-\frac{3}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{5}{13}\right)\left(\frac{4}{5}\right)  \;=\;\boxed{-\frac{56}{65}}

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