SOLUTION: Given sin (A) = 3/5 with 0 < A < 90 and cos (B) = 4/5 with 270 < B < 360, which expression gives the value of sin(A-B)? A. -24/25 B. 0 C. 24/25 D. 1

Algebra ->  Trigonometry-basics -> SOLUTION: Given sin (A) = 3/5 with 0 < A < 90 and cos (B) = 4/5 with 270 < B < 360, which expression gives the value of sin(A-B)? A. -24/25 B. 0 C. 24/25 D. 1      Log On


   

Question 1114968: Given sin (A) = 3/5 with 0 < A < 90 and cos (B) = 4/5 with 270 < B < 360, which expression gives the value of sin(A-B)?
A. -24/25
B. 0
C. 24/25
D. 1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Using the pythagorean identity, you'll find that
cos(A) = 4/5
sin(B) = -3/5
Note: Since B is in quadrant 4, the sine value here is negative

So,
sin(A-B) = sin(A)*cos(B)-cos(A)*sin(B)
sin(A-B) = (3/5)*(4/5) - (4/5)*(-3/5)
sin(A-B) = 12/25 + 12/25
sin(A-B) = 24/25