三角函数的问题,急200分有点英文我也不知道中文是什么,相信应该会看的懂吧?一定要有过程,3.f(Ø)=c
5个回答

f(Ø)=cot(πØ+2π/3)

T=π/w

=π/π

=1

φ=2π/3

定义域 πØ+2π/3≠π+kπ k∈Z

πØ≠π/3+kπ

Ø≠k+1/3

值域为R

令Ø=0

cot(2π/3)

=- √3

令y=0 cot(πØ+2π/3) =0

πØ+2π/3= kπ k∈Z

πØ= kπ-2π/3

Ø= k-2/3

渐近线的方程 πØ+2π/3=π+kπ k∈Z

πØ=π/3+kπ

Ø=k+1/3

2.f(Ø)= -2cos(2Ø-2/3)

A=2

T=2π/W

=2π/2

=3π

φ=2/3

定义域为R

值域 [-2,2] 令y=0 2sin(2Ø-2/3)=0

sin(2Ø-2/3)=0

2Ø-2/3=kπ k∈Z}

2Ø= kπ+2/3

Ø= kπ/2+1/3

令Ø =0 y=2sin(2Ø-2/3)

= sin2/3