If sin a = 3/5 when Pi/2<a<Pi and
cos B = -12/13 when Pi<B<3Pi/2, then what is the value of sin(a+b)?
I don't get the question... Help appreciated.
pre-calculus - Reiny, Thursday, January 28, 2010 at 10:27pm
make two separate diagrams,
one will be the 3,4,5 right-angled triangle,
the other will be the 5,12,13 right-angled triangle
given sin a = 3/5 and a is in the II quadrant, so cos a = -4/5
given cos B = -12/13 and B is in III, so
sin B = - 5/13
now you have to know the expansion for
sin(a+b) = sina cosb + cosa sinb
=(3/5)(-12/13) + (-4/5)(-5/13) = -16/65
pre-calculus - Kimmie, Thursday, January 28, 2010 at 10:57pm