SOLUTION: Given sinA=6/7 and cosB=-1/6, where Ais in quadrant II and B is in quadrant III, find sin(A+B), cos(A-B), and tan(A-b)

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Question 1042627: Given sinA=6/7 and cosB=-1/6, where Ais in quadrant II and B is in quadrant III, find sin(A+B), cos(A-B), and tan(A-b)
Answer by ikleyn(52879) About Me  (Show Source):
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Given sin(A) = 6/7 and cos(B) = -1/6, where A is in quadrant II and B is in quadrant III, find sin(A+B), cos(A-B), and tan(A-b)
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Similar problems were solved many times here.
See the lessons
    - Calculating trigonometric functions of angles
    - Advanced problems on calculating trigonometric functions of angles
in this site. Find appropriate samples for you there. 

The instructions are as follows:

1.  Knowing that sin(A) = 6%2F7, calculate cos(A) = +/- sqrt%281-sin%5E2%28A%29%29.

2.  Knowing that the angle A is in QII, determine the sign at the square root in the n.1 above.

3.  Knowing that cos(B) = -1%2F6, calculate sin(B) = +/- sqrt%281-cos%5E2%28B%29%29.

4.  Knowing that the angle B is in QIII, determine the sign at the square root in the n.3 above.


5.  As a last step, apply the formulas 

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

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

       tan(A-B) = %28tan%28A%29+%2B+tan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29.


Notice that you just have everything to substitute into these formulas.

See also the lessons
    - Addition and subtraction formulas
     - Addition and subtraction formulas - Examples
in this site.