cos2x (sinx cosx)积分

解（1）f(x)＝32sin2x+1+cos2x2+a＝sin(2x+π6)+a+12，（2分）∴T=π．（4分）由π2+2kπ≤2x+π6≤3π2+2kπ，得π6+kx≤x≤2π3+kπ．故函数f（

解（1）f(x)＝32sin2x+1+cos2x2+a＝sin(2x+π6)+a+12，（2分）∴T=π．（4分）由π2+2kπ≤2x+π6≤3π2+2kπ，得π6+kx≤x≤2π3+kπ．故函数f（

原式=√3/2*sin2x+(1+cos2x)/2=√3/2*sin2x+1/2*cos2x+1/2=sin2xcosπ/6+cos2xsinπ/6+1/2=sin(2x+π/6)+1/2

sinx/cosx=tanx=2sinx=2cosx带入恒等式sin²x+cos²x=1cos²x=1/5sinxcosx=(2cosx)cosx=2cos²x

f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/

f(x)=2√3sinxcosx-cos2x=√3sin2x-cos2x=2(sin2x*√3/2-cos2x*1/2)=2sin(2x-π/6)x=π/12;函数f(x)的图象可以由函数y(x)=2

Letu=1+sin(x)cos(x)=1+(1/2)sin(2x)anddu=cos(2x)dx→dx=du/cos(2x)So∫cos(2x)/(1+sin(x)cos(x))dx=∫1/udu=

（I）由题意得，f(x)＝32sin2x−1+cos2x2+12=32sin2x−12cos2x=sin(2x−π6)，∴函数f（x）的最小正周期T=2π2=π，（II）列表如下：xπ12π37π12

由题意知，k=cos2x-23sinxcosx-1=cos2x-3sin2x-1=2cos（2x+π6）-1当x∈R时，cos（2x+π6）∈[-1，1]∴2cos（2x+π6）∈[-2，2]∴2co

分两部分求2sin2x=4sinxcosx注：sin2x=2sinxcosx=4sinxcosx/{(cosx)^2+(sinx)^2}注：{(cosx)^2+(sinx)^2=1=4tanx/{1+

tanx=3cosx^2=1/(1+tanx^2)=1/10sinxcosx+cos2x=tanx*cosx^2+2cosx^2-1=3/10+2/10-1=-1/2

f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)2x+π/4=π/2+2kπ时f(x)有最大值f(x)=√2x=π/8+kπ2x+π/4=3π/2+2kπ时

f(x)=sin2x+cos2x=√2sin(2x+π/4)f(π/4)=√2sin(2*π/4+π/4)=√2*√2/2=10

f(x)=sin2x+cos2x=√2sin(2x+π/4)所以T=2π/2=π最大值=√2f(θ+π/8)=√2sin(2θ+π/4+π/4)=√2cos2θ=√2/3cos2θ=1/3θ锐角则si

f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1

cos2x-2根号3sinxcosx=cos2x-根号3（2sinxcosx）运用倍角公式得=cos2x-根号3sin2x运用辅助角公式得=-2sin(2x-六分之π)由ω=2,T=二派除以ω,所以周

解:原式=√3sin2x＋cos2x＋1=2(√3/2sin2x＋1/2cos2x＋1=2cos(2x-pai/3)＋1.

∫（COS2X）／（1十SinXCOSX）dX=∫(1/2)/(1+sin2x/2)d(sin2x)=∫(1/2)/(1+u/2)du(u=sin2x)=∫1/(u+2)d(u+2)=ln|u+2|+

不对,因为f（x）=cos2x-2√3sinxcosx=cos2x-√3sin2x=2[sinπ/6cos2x-cosπ/6sin2x]=2sin(π/6-2x)左移5π/12,f（x）=2sin【π

⑴接着你做的,sin(π/2)+cos(π/2)=1+0=1⑵f(α/2)=sinα+cosα=√2/2两边平方得1+2sinαcosα=1/2sin2α=-1/22α=150°,α=75°sinα=