SOLUTION: Given that cotx=-12/5 with x in the IV quadrant and secy=-17/15 with y in the III quadrant find csc(x+y)? I have found everything else for the triangle but I just do not know which

Algebra ->  Trigonometry-basics -> SOLUTION: Given that cotx=-12/5 with x in the IV quadrant and secy=-17/15 with y in the III quadrant find csc(x+y)? I have found everything else for the triangle but I just do not know which      Log On


   

Question 816073: Given that cotx=-12/5 with x in the IV quadrant and secy=-17/15 with y in the III quadrant find csc(x+y)? I have found everything else for the triangle but I just do not know which identity to use. I know that the csc is 1/sin so do I use the sin(u+v) identity?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Given that cotx=-12/5 with x in the IV quadrant and secy=-17/15 with y in the III quadrant find csc(x+y)?
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csc(x+y)=1/sin(x+y)
The problem then is to solve for sin(x+y)then take its reciprocal.
sin(x+y)=sinxcosy+cosxsiny
hypotenuse of reference right triangle in quadrant IV=√(12^2+5^2)=√144+25)=√169=13
cotx=-12/5 (given)
sinx=-5/13
cosx=12/13
..
In quadrant III:
secy=-17/15 (given)
siny=-15/17
cosy=-√(1-sin^2y)=-√(1-(15/17)^2)=-√(1-225/289)=-√(64/289)=-8/17
..
sin(x+y)=(-5/13)*(-8/17)+(12/13)*(-15/17)
=40/221-180/221=-140/221
csc(x+y)=-221/140
..
calculator check:
tanx=-5/12
x=337.38˚
..
cosy=8/17
y=241.93˚
..
x+y=579.31˚
..
sin(x+y)=sin(579.31˚)≈-0.6335
Exact value as calculated=-140/221≈-0.6335
..