Question 602906
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A is an angle and they tell you it is in the first quadrant and that the tangent is *[tex \LARGE \frac{9}{40}]


B is another angle.  You are given that this angle is in the third quadrant and that the cosine is *[tex \LARGE -\frac{4}{5}]


They want to subtract the measure of angle B from the measure of angle A and then take the sine of the difference.


For such an operation you need the difference formula for sine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A\ -\ B)\ =\ \sin(A)\cos(A)\ -\ \cos(A)\sin(B)]


Remember that tangent is sine over cosine, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin(A)}{\cos(A)}\ =\ \frac{9}{40}]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 40\sin(A)\ =\ 9\cos(A)]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1600\sin^2(A)\ =\ 81\cos^2(A)] 


Use the Pythagorean Identity and substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1600\(1\ -\ cos^2(A)\ =\ 81\cos^2(A)]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(A)\ =\ \frac{1600}{1681}]


Take the positive (remember, QI) square root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(A)\ =\ \frac{40}{41}]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(A)\ =\ \frac{81}{1681}]


Again, the positive square root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A) =\ \frac{9}{41}]


Similarly you can get from *[tex \LARGE \cos(B)\ =\ -\frac{3}{5}] to *[tex \LARGE \sin(B)\ =\ -\frac{4}{5}].  QIII, both sine and cosine are negative.


Now you have the four values you need to plug into:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A\ -\ B)\ =\ \sin(A)\cos(A)\ -\ \cos(A)\sin(B)]


You can do your own arithmetic.


Then, for futher study, here are the other three formulas you will need:


<a href="http://oakroadsystems.com/twt/sumdiff.htm">Sum and Difference Formulas</a>


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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