SOLUTION: Given sinA = 5/13 and cosB = 4/5, then to the nearest hundredth, sin(B-A)= The only one's I know of are sin(A+B) and sin(A-B) so I don't know how to approach this question. Give

Algebra ->  Trigonometry-basics -> SOLUTION: Given sinA = 5/13 and cosB = 4/5, then to the nearest hundredth, sin(B-A)= The only one's I know of are sin(A+B) and sin(A-B) so I don't know how to approach this question. Give      Log On


   

Question 1040904: Given sinA = 5/13 and cosB = 4/5, then to the nearest hundredth, sin(B-A)=
The only one's I know of are sin(A+B) and sin(A-B) so I don't know how to approach this question. Given the way the other two are set up I would guess that it would be sin(B-A)= sinB cosA - cosB sinA but I'm not totally sure
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
How about this?
sin%28x-y%29=sin%28x%29cos%28y%29-cos%28x%29sin%28y%29
So in this case x=B and y=A.
So,
sin%28y%29=5%2F13
and
cos%28y%29=12%2F13 or cos%28y%29=-12%2F13
and
cos%28x%29=4%2F5
and
sin%28x%29=3%2F5 or sin%28x%29=-3%2F5
Take it from there and complete it.
Don't get hung up on the variables, they're just placeholders.
When you run into something like this where it's confusing, just change the variable names and keep your head straight.
Good luck!