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y=cosx y∈<-1;1>
y=cos^2x y∈<0;1>
f(x)=-2cos^2x f(D)=<-2;0>
b) ctgx≠0 i tgx≠0
x≠kπ/2
D=R/(kπ/2)
sinx * cosx/sinx - cosx*sinx/cosx=cosx-sinx= √2(√2/2cosx-√2/2sinx)
√2/2 = sinπ/4 i cosπ/4
zatem korzystamy z wzoru na różnicę kątów
√2(sinπ/4cosx-cosπ/4sinx)=√2sin(π/4-x)=√2sin(-(x-π/4)=
sin-x=-sinx
=-√2sin(x-π/4)
y=sinx y∈<-1,1>
sin(x-π/4) y∈<-1;1>
f(x)=-√2sin(x-π/4)=<-√2,√2> dla x ∈D