tan(b/2)=1/2,sin(a+b)=5/13,a属于(0,π),b属于(0,2π)
(1)求sinb,cosb
(2)求sina
万能公式
(1) sinb=2*tan(b/2) / ( 1 + (tan(b/2))^2 ) = 1/(1+1/4)=4/5
cosb=( 1-(tan(b/2))^2 ) / ( 1 + (tan(b/2))^2 ) = (1-1/4)/(1+1/4)=3/5
(2) 由于 sinb>0,cosb>0,可知b属于(0,π/2),由题设a属于(0,π),所以a+b属于(0,3π/2),但是由于sin(a+b)=5/13>0,所以a+b属于(0,π),
如果a+b属于(0,π/2),那么 π/2>a+b>a>0,应该有sin(a+b)=5/13 > sin(a)=4/5,但这是不成立的,故a+b属于(π/2,π),因此
cos(a+b)=-sqrt(1-(5/13)^2)=-12/13.
故sina=sin((a+b)-b)=sin(a+b)cosb-cos(a+b)sinb
=5/13 * 3/5 -(-12/13)* 4/5 = 63/65.