How to find sin(α-β)=

Question

I was given this problem and don't know how to solve it.

Answer 1

Isn't there an error in the question?

if angle B is in 3rd quadrant, then cos(B)<0

say cos(B)=-12/13 in third quadrant sin(B)=±(1-cos²(B))^(1/2)

sin(B)=-5/13 since B in 3rd quadrant.

sin(A)=3/5,,,A in 1st quadrant

cos(A)=±(1-sin²(A))^(1/2)

cos(A)=+4/5 since A in 1st quadrant.

now sin(A-B)=(3/5)(-12/13)-(-5/13)(4/5)=sinAcosB-sinBcosA

=-16/65

sin^-1(-16/65)= -.25 radian (4th quadrant)

but quadrant 1 angle minus quadrant 3 angle must be must be quadrant 3 or 2.

since sin(A-B)<0,,quadrant 3

sin^-1(sin(A-B)=3.14+.25=3.39 radians.

yay sin(3.39)=-.25

Answer 2

Recall sin(α-β) = sin(α)cos(β) - cos(α)sin(β)

Because sin(α) = 3/5 = opposite/hypotenuse, then cos(α) = 4/5 = adjacent/hypotenuse

Because cos(β) = 12/13 = adjacent/hypotenuse, then sin(β) = 5/13 = opposite/hypotenuse

Hence, sin(α-β) = (3/5)(12/13) - (4/5)(5/13) = 36/65 - 20/65 = 16/65

Hope this helped!