Given csc A=-13/12 and cot B= 4/3 3pi/2<A<2pi and pi<B<3pi/2. What is Sin (A+B) and Cos (A-B)

Answer 1

cscA = -13/12 with A in the 4th quadrant

cotB = 4/3 with B in the 3rd quadrant

cscA = 1/inA = -13/12

sinA =-12/13

cosA = 5/13

It's a 5-12-13 right triangle

A = -67.4

cotB = 4/3 in Quadrant III implies cosB = -4/5

cotB = 4/3 implies cosB = -4/(sqr(3^2 + 4^2) =-4/sq25 = -4/5

cosB=-4/5

sinB = -3/5

It's a 3-4-5 right triangle

B=216.9 degrees, (180+36.9) using inverse trig function on a calculator

A+B = -67.4 + 216.9 = 149.5 degrees

sin149.5 = -sin31.5 = 0.51

cos149.5 =-cos31.5= -0.86

sin(A+B) = sinAcosB + cosAsinB = -12/13(-4/5) + 5/13(-3/5)=48/65-15/65=33/65=approximately 0.51

cos(A+B) = cosAcosB-sinAsinB = (5/13)(-4/5)-(-12/13)(-3/5) = -20/65-36/65 = -56/65=-approximately -0.86

once we have sin(A+B)=33/65

cos(A+B) = -sqr(65^2-33^2)/65 = -sqr[4225-1089)/65 = -sqr3136/65 = -56/65 = -0.86

It's easy to get the signs wrong. Just check at the end to make sure the sign fits the quadrant

In the second quadrant, where A+B is, sines > 0 and cosines <0

Answer 2

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