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May 12th, 2008, 01:37 PM   #1
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finding values for trigonometric expression

Can someone help me with this question?

Determine the exact value for each trigonometric expression, given sinA= 4/5, cos B= (-8/17),
[(pi)/2] < A < pi and (pi) < B < [3(pi)/2]

Find
1) sin (A-B)

2) Sin (A+B)

3) cos (A+B)

4) tan (A-B)

if you could just tell me one of the answers and how you got to it, or even just help me get it started, i can probably figure out the rest. I just dont even know where to begin! thank youuuuuu!!
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May 12th, 2008, 01:39 PM   #2
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correction

all the < signs should be "less than or equal to" ..sorry!
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May 13th, 2008, 02:14 PM   #3
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You can use "<" instead of "≤", as the values given for sin(A) and cos(B) are inconsistent with equality.

sin(pi - A) = sin(A) = 4/5,
so cos(A) = -cos(pi - A) = -3/5 (by Pythagoras's theorem).

cos(B - pi) = -cos(B) = 8/17,
so sin(B) = -sin(B - pi) = -15/17 (by Pythagoras's theorem).

1) sin(A - B) ≡ sin(A)cos(B) - cos(A)sin(B) = -(4/5)(8/17) - (3/5)(15/17) = -77/85.

2) sin(A + B) ≡ sin(A)cos(B) + cos(A)sin(B) = -(4/5)(8/17) + (3/5)(15/17) = 13/85.

(This is trigonometry, not calculus, so the topic will probably be moved from this section after a while.)

3) cos(A + B) ≡ cos(A)cos(B) - sin(A)sin(B) = (3/5)(8/17) + (4/5)(15/17) = 84/85.

4) tan(A - B) = sin(A - B)/cos(A - B) = (-77/85)/(-36/85) = 77/36.
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