1 Given that cos(A-B) / cos(A+B) = 7/3, and that A is acute, tan B = 2, find
a. sin A
2 Given that A = 3/5, cos B = 1/ sq.rt 2 and that A and B are acute, find the value of sin(A+B) and cos (A-B)...
just wondering, is there anything wrong with this question?
1. (cosAcosB+sinAsinB)/(cosAcosB-sinAsinB)=7/3
3cosAcosB+3sinAsinB = 7cosAcosB-7sinAsinB
10sinAsinB=4cosAcosB
10sinA/cosB = 4cosB/sinB
10tanA = 4(1/2)
tanA = 1/5
Since tan A = 1/5 construct a triangle in 1st quadrant. A=1 and O =5, thus H=sqrt(26)
sinA = O/H
sin A = 5/(sqrt26)
2. Should be something wrong. should be sin/cos/tan/whatever A = 3/5 normally.