Question 817936: How do I find sin (x-y) and tan (x+y) if cos x=(-1/3) and tan y= (1/2) with x being a quadrant II angle and y a quadrant III angle? Thank you for your help!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! How do I find sin (x-y) and tan (x+y) if cos x=(-1/3) and tan y= (1/2) with x being a quadrant II angle and y a quadrant III angle?
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Identity: sin(x-y)=sinxcosy-cosxsiny
Identity: tan(x+y)=(tanx+tany)/(1-tanxtany)
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cosx=(-1/3)(In quadrant II where cos<0, sin>0, tan<0)
sinx=√(1-cos^2x)=√(1-1/9)=√(8/9)=√8/3
tanx=sinx/cosx=√8/-1=-√8
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tany=1/2(In quadrant III where sin<0, cos<0, tan>0)
hypotenuse of reference right triangle in quadrant III=√((1^2)+(2^2))=√(1+4)=√5
siny=-1/√5=-√5/5
cosy=-2/√5=-2√5/5
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sin(x-y)=√8/3*-2√5/5-(-1/3)*-√5/5=-(2√40+√5)/15
tan(x+y)=(-√8+1/2)/(1-(-√8)*1/2=(-√8+1/2)/(1+√8/2)=((-2√8+1)/2)/((2+√8)/2)
=(-2√8+1)/(2+√8)
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Calculator check:
cosx=-1/3
x≈109.47˚(Q2)
tany=1/2
y≈206.57˚(Q3)
x+y≈316.04˚
x-y≈-97.1˚
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sin(x-y)≈sin(-97.1)≈-0.9923..
exact value as calculated=-(2√40+√5)/15≈-0.9923..
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tan(x+y)≈tan(316.04)≈-0.964..
exact value as calculated=(-2√8+1)/(2+√8)≈-0.964..
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