已知5cosa+12sina可表示为Asin（a+b）的形式，其中A>0,0《b<2派,求A及cos2b的值

5^2+12^2=13^2
5cosa+12sina
=13*(5/13cosa+12/13sina)
设sinx=5/13,cosx=12/13.(0〈x<派/2)。得
5cosa+12sina
=13*(5/13cosa+12/13sina)
=13*（sinxcosa+cosxsina)
=13sin(a+x)
A=13，
b=arcsin5/13.
sinb=5/13,
cos2b=1-2sinb^2=1-2*(5/13)^2=119/169

13（5/13 cosa + 12/13 sina )
令 b = arctan 5/12
则为 13 sin(a+b)
且 b=arccos 12/13
cos2b= 2(cos b)2 -1
=(288-169)/169
=119/169