Find a, b, c,d,e & f.?

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1 Answer
Jun 24, 2018

Use standard formulae to classify the situation

Explanation:

Recall angle addition formulae:
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-cosAsinB
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
tan(A-B)=(tanA-tanB)/(1+tanAtanB)

Recall also the Pythagorean identity:
sin^2x+cos^2x=1
which tells us
|sinA|=15/17
|cosB|=3/5

The given angular interval for both A and B, ((3pi)/2,2pi), corresponds to the fourth quadrant of the Cartesian plane. We use this to deduce the signs of sinA and cosB - sine is negative in this quadrant and cosine positive. So
sinA=-15/17
cosB=3/5
and thus
tanA=-15/8
tanB=-4/3

We now have all the knowledge we need to answer the specific questions:

a)
sin(A+B)=sinAcosB+cosAsinB=
-15/17*3/5+8/17*(-4/5)=-77/85

b)
sin(A-B)=sinAcosB-cosAsinB=
-15/17*3/5-8/17*(-4/5)=-13/85

c)
tan(A+B)=(tanA+tanB)/(1-tanAtanB)=
(-15/8-4/3)/(1-(-15/8)*(-4/3))=(-77/24)/(-36/24)=77/36

d)
tan(A-B)=(tanA-tanB)/(1+tanAtanB)=
(-15/8-(-4/3))/(1+(-15/8)*(-4/3))=(-13/24)/(84/24)=-13/84

e)
sin(A+B) is negative, so we are either in quadrant 3 or 4.
tan(A+B) is positive, so we are either in quadrant 1 or 3.
Therefore A+B is in quadrant 3.

f)
sin(A-B) is negative, so we are either in quadrant 3 or 4.
tan(A-B) is negative, so we are either in quadrant 2 or 4.
Therefore A-B is in quadrant 4.