F(x)= vector Mn-1/2 = sinwxcoswx+(coswx) 2-1/2 =1/2sin2wx+1/2cos2wx = √ 2/2sin (2wx+)

The minimum positive period is 4, w = 1/4, and f (x) = √ 2/2 sin (1/2x+π/4).

(1) When sin( 1/2x+π/4)=- 1, the minimum value is -√2/2. At this time, 1/2x+π/4 =-π/2+2kπ, and x =-2kπ.

When sin( 1/2x+π/4)= 1, the maximum value is √2/2. At this time, 1/2x+π/4 = π/2+2kπ. X = π/2+4kπ。

(2)(2a-c)cosB=bcosC

According to sine theorem a/sinA=b/sinB=csinC=2R.

Then (2sinA-sinC)cosB=sinBcosC.

2sinAcosB-sinCcosB=sinBcosC

2sinAcosB=sinCcosB+sinBcosC

2sinAcosB=sin(B+C)

2sinAcosB=sin( 180-A)

2sinAcosB=sinA

Sina ≠0, so cosB= 1/2, B=π/3.

A=2π/3-C，0，4， 1

f(x)= 2 sinwxcoswx- 1/2 = sin2wx- 1/2

t = 2π/2w = 4π2w = 1/2f(x)= sin(x/2)- 1/2

X=π+4kπ(k integer), f (x) max =1-1/2 =1/2.

X=3π+4kπ(k integer) f(x) Minimum value =- 1- 1/2=-3/2.

2

a/sinA=b/sinB=c/sinC=k

(2a-c) cosb = bco...，2，f(x)= sinwxcoswx+(coswx)2- 1/2 = 1/2 * sin2wx+ 1/2 * cos2wx-65438。 f(x)=√2/2 * sin( 1/2 * x+π/4)- 1

(1) When sin (1/2 * x+π/4) =-1is the minimum, at this time, 1/2*x+π/4=-π/2+2kπ, and x =-3π.

0, a high school math trigonometric function problem …

M = (sinwx，coswx)，n = (coswx，coswx)，(w >； 0) If the minimum positive period of the function f(x)=m*n- 1/2 is 4.

(1) Find the maximum value of the function * * * (2) (2a-c) COSB = BCOSC, and find the value range of the function f(A).