Question

高一数学。 手写过程

Answer 1

1、因为：sin2x=2sinx *cosx cos2x=(cosx)^2-(sinx)^2
f(x)=asinx-cosx
a=1
f(x)=sinx-cosx=√2sin(x-π/4)
最小正周期 x-π/4=2π x=9π/4
对称轴为x=π/4
2、a=2
f（x）=2sinx-cosx=0
2sinx=cosx
cos2x/(1+sin2x)=[(cosx)^2-(sinx)^2]/[(sinx)^2+2sinxcosx+(cosx)^2]
=(cosx+sinx)(cosx-sinx)/(sinx+cosx)^2
=(cosx-sinx)/(sinx+cosx)
=sinx/(3sinx)=1/3

Answer 2

1
a=1
f(x)=sinx-cosx=√2[sin(x-π/4)]
T=2π
x=kπ+9π/4
2
a=2
f(x)=2sinx-cosx=0
tanx=1/2
cos2x/[1+sin2x]
=[cox^2(x)-sin^2(x)]/[1+2sinxcosx]
=[1-tan^2(x)]/[1/cos^2(x)+2/tanx]
=(1-1/4)/[sec^2(x)+1]
=(3/4)/[1+tan^2(x)+1]
=(3/4)/[2+1/4)
=1/3