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?
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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=&#8730;(1-cos^2x)=&#8730;(1-1/9)=&#8730;(8/9)=&#8730;8/3
tanx=sinx/cosx=&#8730;8/-1=-&#8730;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=&#8730;((1^2)+(2^2))=&#8730;(1+4)=&#8730;5
siny=-1/&#8730;5=-&#8730;5/5
cosy=-2/&#8730;5=-2&#8730;5/5
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sin(x-y)=&#8730;8/3*-2&#8730;5/5-(-1/3)*-&#8730;5/5=-(2&#8730;40+&#8730;5)/15
tan(x+y)=(-&#8730;8+1/2)/(1-(-&#8730;8)*1/2=(-&#8730;8+1/2)/(1+&#8730;8/2)=((-2&#8730;8+1)/2)/((2+&#8730;8)/2)
=(-2&#8730;8+1)/(2+&#8730;8)
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Calculator check:
cosx=-1/3
x&#8776;109.47&#730;(Q2)
tany=1/2
y&#8776;206.57&#730;(Q3)
x+y&#8776;316.04&#730;
x-y&#8776;-97.1&#730;
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sin(x-y)&#8776;sin(-97.1)&#8776;-0.9923..
exact value as calculated=-(2&#8730;40+&#8730;5)/15&#8776;-0.9923..
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tan(x+y)&#8776;tan(316.04)&#8776;-0.964..
exact value as calculated=(-2&#8730;8+1)/(2+&#8730;8)&#8776;-0.964..