∵ This function is odd function, ∴ f (-x) =-f (x) =-cos (x)-kcos (x-π/3).
∴cosx+kcos(x+π/3)=-cos(x)-kcos(x-π/3)
∴2 cos x+k(cos(x+π/3)+cos(x-π/3))= 0
According to the trigonometric function sum difference formula, there are:
2 cosx+k(cosx cos(π/3)-sinx sin(π/3)+cosx cos(π/3)+sinx sin(π/3))= 0
2cosx+2kcosxcos(π/3)=0
k =- 1/cos(π/3)=- 1/( 1/2)=-2
(2)f(x)= cosx-2 cos(x-π/3)= cosx-2(cosx cos(π/3)+sinx sin(π/3))
=cosx-2(cosx/2+√3/2*sinx)
=-√3sinx
According to the property of trigonometric function y=sinx, x is a increasing function in 2kπ-π/2, 2kπ+π/2, and a decreasing function (k∈z) in 2kπ+π/2, and there are:
In the range of 2kπ-π/2 and 2kπ+π/2, x2>x 1, sinx2 & gtsinx 1, sinx2-sinx1>; 0
∴-√3(sin x2-sinx 1)& lt； 0，-√3(sin x2)& lt； -√3(sinx 1)，f(x2)& lt； f(x 1)
∴ f(x)=cosx-2cos(x-π/3), which is a decreasing function in the range of 2kπ-π/2 and 2kπ+π/2.
Similarly, f(x)=cosx-2cos(x-π/3), and x is increasing function in 2kπ+π/2 and 2kπ+3π/2.
2. Solution: (1) f (x) = (cosx-sinx)/cosx * (1+√ 2 (sin2x * cosπ/4+cos2x * sinπ/4))
=(cosx-sinx)/cosx *( 1+sin2x+cos2x)
=(cosx-sinx)/cosx*( 1+2sinxcosx+ 1-2(sinx)^2)
=(cosx-sinx)/cosx*(2-2(sinx)^2+2sinxcosx)
=(cosx-sinx)/cosx*(2(cosx)^2+2sinxcosx)
=(cosx-sinx)/cosx * 2 cosx(cosx sinx)
=2 (cosine-sine) (cosine+sine)
=2((cosx)^2-(sinx)^2)
=2cos2x
Because tanx, x≠kπ+π/2, the domain of this function is x ≠ kπ+π/2 (k ∈ z);
Because-1≤cos2x≤ 1, the range of f(x) is-2 ≤ 2cos2x ≤ 2;
(2) Because f(x)=2cos2x and the period of this function is 2π/2=π, then:
F(x) is a increasing function in the range of kπ-π/2 and kπ.
F(x) is a increasing function in the range of kπ, kπ+π/2.
3. Solution: (1) Because √2sin(x-π/4)≠0, so x≠2kπ+π/4, then
The domain of this function is x ≠ 2kπ+π/4 (k ∈ z);
(2) cos2x = cosx 2-sinx 2 = (cosx+sinx) (Coase-Sinks)
√2 sin(x-π/4)=√2(sinx cos(π/4)-cosx sin(π/4))= sinx-cosx =-(cosx-sinx)
f(x)= 1-(cosx+sinx)(cosx-sinx)/(-(cosx-sinx))= 1+cosx+sinx
= 1+√2(sinx cos(π/4)-cosx sin(π/4)))
= 1+√2sin(x+π/4)
Because, y=sinx, increasing function (k∈z) in 2kπ-π/2, 2kπ+π/2, and the decreasing function in 2kπ+π/2, and the period of this function is 2π; X+π/4=-π/2, x=-3π/4, x+π/4=π/2, x=π/2, x=5π/4, so the monotonic interval of this function is:
F(x) is increasing function in the range of 2kπ-3π/4 and 2kπ+π/2.
F(x) is a decreasing function in the range of 2kπ+π/2 and 2kπ+5π/4.