Posted by shay on .
Given cos a=7/25 in quadrant II , and sin b=12/13, in quadrant IV , find cos(a+b) and sin2a

mth, trig 
Steve,
sin a = 24/25
cos b = 5/13
now apply addition formulas 
mth, trig 
shay,
Thanks but how did you find that? My numbers came out completely different _

mth, trig 
Steve,
If a is in QII, sin is positive, cos is negative.
7^2 + 24^2 = 25^2, so it's a 72425 triangle
If b is in QIV, sin is negative, cos is positive.
5^2 + 12^2 = 13^2, so it's a 51213 triangle.
what did you do to find the missing sin/cos values?