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    If sin (A) = 4/5 and cos (B) = 12/13 where A is obtuse and B is acute, find the value of sin(A-B)?

    My working

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

    (4/5) x (5/13) - (3/5) x (12/13) (triangles)

    20/65 - 36/65 = -16/65

    3/5 should be a minus? Not sure if it is or why it is
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    (Original post by H3nners)
    If sin (A) = 4/5 and cos (B) = 12/13 where A is obtuse and B is acute, find the value of sin(A-B)?

    My working

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

    (4/5) x (5/13) - (3/5) x (12/13) (triangles)

    20/65 - 36/65 = -16/65

    3/5 should be a minus? Not sure if it is or why it is
    It should indeed be negative.

    Consider the graphs of \sin(x) (red) and \cos(x) (green) where the angle A is obtuse, meaning it is 90<x<180 degrees which is the region I have marked blue on the diagram below:
    Name:  Capture.PNG
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Size:  17.2 KB

    Now sine of an obtuse angle is clearly positive as that portion of \sin(x) us always above 0, however cosine of an obtuse angle is clearly negative as that portion of \cos(x) is below 0.

    So, since A is obtuse, \sin(A)>0 and \cos(A)<0

    P.S. Check your compound angle identity that you've used as well, it's not the right one.
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    (Original post by H3nners)
    Sin(A-B)= sin(A) x sin(B) - cos(A) x cos(B)

    (4/5) x (5/13) - (3/5) x (12/13) (triangles)

    20/65 - 36/65 = -16/65

    3/5 should be a minus? Not sure if it is or why it is
    your expansion is wrong

    it should say

    Sin(A-B)= sin(A) x cos(B) - cos(A) x sin(B)
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    Yeah my mistake I typed in the formula wrong sorry about that. Cheers rdk makes a load more sense will draw the graphs from now on. Sick explanation!
 
 
 
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