1.已知函数发f(x)=asinx+btanx+1满足f(5)=7,则f(-5)的值为( ）2．f(cosx=cos3x,那么f(sin30度）的值为（ ）

∵函数f (x )=asinx+btanx+1
∴f(5)=asin5+btan5+1=7
∴asin5+btan5=7-1=6
∵f(-5)=-asin5-btan5+1
∴f(-5)=-(asin5+btan5)+1=-6+1=-5
2)因为sin30°＝cos60°，
故f(sin30°)=f(cos60°)=cos(3*60°)=cos180°=-1

1.因为f(x)=asinx+btanx+1所以f(-x)=-asinx-btanx+1即f(x)+f(-x)=2
则f(-5)=2-f(5)=-5
2.f(sin30度）=f(cos60度)=cos180度=-1

1.f(x)=asinx+btanx+1
=> f(x) - 1 = asinx + btanx
令h（x）= f(x) - 1
h(-x) = a(sin(-x))+ b(tan(-x))
= -asinx - btanx
= - h(x)
=> h(-5)= f(-5) - 1 (1)
h(-5)= -h(5) = - (f(5)-1) = -(7 - 1)= -6 (2)
(1)(2) => f(-5) = -5
2.f(sin30度) = f(cos60度)= cos （3*60度）=cos 180度 = -1