Cos A =4/5(A in quadrant 4) and Sin B=2/3 (B in quadrant 2)

Question

Using the equation above I need to find

Sin B-A and cos B/2

Answer 1

cosA = 4/5 A in quadrant IV

opposite side = -sqr(5^2-4^2) = -sqr(25-16)=-sqr9= -3

sinA = opposite side/hypotenuse = -3/5

sinB = 2/3 B in quadrant II

adjacent side = -sqr(3^2-2^2) =-sqr(9-4) = -sqr5

cosB = adjacent side/hypotenuse = -sqr5/3

sin(B-A) = sinBcosA - sinAcosB = (2/3)(4/5) -(-3/5)(-sqr5/3) = 8/15 -3sqr5/15 = (8-3sqr5)/15 = about (8-6.71)/15 = 1.29/15 = 0.09

check the solution, with a trig calculator with inverse function

A= approximately -36.87

B = 180 -41.81 = approximately 138.19

B-A = 138.19--36.87 = approximately 175.06

sin175.06 = about 0.09

cos(B/2) = about cos(138.19/2) = -0.357

cos(B/2) = -sqr((1+cosB)/2)=-sqr((1-sqr5/3)/2)

= about -0.357