Trigonometry Question

Question

given sin(a)=4/5 and cos(b)=1/3, with a and b both in interval [pi/2,pi): find sin(a-b) and find cos(a+b)

Answer 1

quadrant 2, sine is positive and cosine is negative

sinA = 4/5 ---> opposite is 4 and hypotenuse is 5 --> adjacent = -3 by pythagorean

sin A = 4/5 and cos A = -3/5

cos b = -1/3 ---> adjacent is -1 and hypotenuse is 3

then opposite is 2 * sqrt(2) by pythagorean

so then sin b = 2*sqrt(2)/3 = (2/3) sqrt(2)

sin(A+B) = sin A cos B + sin B cos A = (4/5)(-1/3) + (2/3) sqrt(2) * (-3/5)

= -4/15 - 6 * sqrt(2) / 5 = (-4-6*sqrt(2))/5

= (-2/5) ( 2 + 3 * sqrt(2))

cos (a+b) = cos a cos b - sin a cos a

= (-3/5)(-1/3) - (2/3)sqrt(2) * (4/5)

= (1/5) - (8/5) sqrt(2)