SOLUTION: Given: A lies in Quadrant III and B lies in Quadrant II,
m1212.
Find the exact value of cos(A - B). SinA= -8/17 CosB= -4/5
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Question 746467: Given: A lies in Quadrant III and B lies in Quadrant II,
m1212.
Find the exact value of cos(A - B). SinA= -8/17 CosB= -4/5
Found 2 solutions by lwsshak3, solver91311:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Given: A lies in Quadrant III and B lies in Quadrant II,
m1212.
Find the exact value of cos(A - B). SinA= -8/17 CosB= -4/5
***
sinA=-8/17
..
cosB= -4/5
sinB=3/5 (from 3-4-5 right triangle)
..
cos(A-B)=cosAcosB+sinAsinB=(-15/17)(-4/5)+(-8/17)(3/5)=60/85-24/85=36/85
..
Check: (with calculator)
sinA=-8/17
A≈208.07º
cosB=-4/5
B≈143.13º
A-B≈64.94º
cos(A-B)≈cos(64.94º)≈0.4235..
Exact ans=36/85≈0.4235..
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ -\frac{8}{17})
\ =\ \frac{64}{289})
\ =\ \frac{64}{289})
\ =\ \frac{225}{289})
\ =\ \pm\frac{15}{17})
But since 
in quadrant III:
\ =\ -\frac{15}{17})
Use the same technique to find )
Then use the formula for the cosine of a difference to calculate the desired value:
\ =\ \cos\alpha\cos\beta\ +\ \sin\alpha\sin\beta)
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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