高職數學求解 幫忙解答一下 謝謝?

1.化簡下列各式：
(1)Tan(－840度)
=Tan(1080度－840度)
=Tan(240度)
=Tan(60度)
=√3
(2)Sin(16π/3)
=Sin(16π/3－4π)
=Sin(4π/3)
=－Sin(π/3)
=－√3/2
2.已知Tan25度=k，試以k表示Sin1285度
Tan25度=k>0
Sin205度=－k/√(1+k^2)<0
Sin1285度
=Sin(1285度－1080度)
=Sin205度
=－k/√(1+k^2)
3.求 Sin(2π－θ)/Cos(π/2+θ)+Tan(－θ)/Tan(π+θ)+Sec(π/2+θ)/Csc(2π+θ)之值
Sin(2π－θ)/Cos(π/2+θ)+Tan(－θ)/Tan(π+θ)+Sec(π/2+θ)/Csc(2π+θ)
=(－Sinθ)/(－Sinθ)+(－Tanθ)/Tanθ+Csc+θ/Cscθ
=1－1+1
=1
4.設0小於等於x小於2π，試求函數f(x)=Cos(2x)－3Sin x+2的最大值
0<=x<2π
f(x)=Cos(2x)－3Sin x+2
=1－2Sin^2 x－3Sinx+2
=－2Sin^2 x－3Sinx+3
=－2(Sin^2 x+3Sinx/2)+3
=－2[Sin^2 x+2*(3Sinx/4)]+3
=－2[Sin^2 x+2*(3Sinx/4)+(3/4)^2]+3+2*9/16
=－2(Sinx+3/4)^2+33/8
－1<=Sinx<=1
－1+3/4<=Sinx+3/4<=1+3/4
－1/4<=Sinx+3/4<=7/4
－1/4<=0<=7/4
(－1/4)^2=1/16，0^2=0，(7/4)^2=49/4
So
0<=(Sinx+3/4)^2<=49/16
0<=2(Sinx+3/4)^2<=49/8
－49/8<=－2(Sinx+3/4)^2<=0
－49/8+33/8<=－2(Sinx+3/4)^2+33/8<=33/8
－16/8<=f(x)<=33/8
－2<=f(x)<=33/8