∫sin2x (1 cos^2x)dx不定积分

解f(x)=(2cos²-1)sin2x+1/2cos4x=cos2xsin2x+1/2cos4x=1/2sin4x+1/2cos4x=√2/2sin(4x+π/4)∴T=2π/4=π/2为

(1)f(x)=cos²xg(x)=1+(1/2)sin2xf(x)=g(x)(cosx)^2=1+(1/2)sin2x(cos2x+1)/2=1+(1/2)sin2xsin2x-cos2x

tanx=tan[(x-π/4)+π/4]=[tan(x-π/4)+tan(π/4)]/[1-tan(x-π/4)tan(π/4)]=(2+1)/(1-2*1)=-3(sin2x+cos2x)/(2c

(1)f(x)=2(cosx)^2+√3sin2x=cos2x+√3sin2x+1=2sin(2x+π/6)+1maxf(x)=3minf(x)=-1(2)f(x)=1-√32sin(2x+π/6)+

y=1/2sin2x+(1+cos2x)/2=1/2*(sin2x+cos2x)+1/2=1/2*√2sin(2x+π/4)+1/2所以值域是[(-√2+1)/2,(√2+1)/2]

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1=-√2(

f(x)=sin2x+2cos²x+1=sin2x+2cos²x-1+2=sin2x+cos2x+2=√2(sin2xcosπ/4+cos2xsinπ/4)+2=√2sin(2x+

f(x)=sin2x+2cos²x+1=sin2x+cos2x+2=√2(√2/2sin2x+√2/2cos2x)+2=√2cos(2x-π/4)+2∴2x-π/4=2kπ时有最大值2+√2

=cosx(1+sinx)再问：��Ϊsin��再答：=1+���(2)*sinx*sin(x+3��/4)再问：������再答：��ô˵��y=cos²x+1/2sin2x=1-(sin

tanx=2所以可得：sin2x=2tanx/(1+tan^2x)=4/5cos2x=(1-tan^2x)/(1+tan^2x)=-3/51+cos^2x=1/2(2cos^2x-1)+1/2+1=1

一样是先化简的问题f(x)=2cos^2x+√3*sin2x-1＝2(cosx)^2-1+√3*sin2x=cos2x+√3*sin2x=2(1/2*cos2x+√3/2*sin2x)=2(sinπ/

∵cos2x=2cos²x-1,∴cos²x=1/2(cos2x+1)∴原式=1/(1/2cos2x+sin2x+1/2)=1/[根号5/2sin(2x+ψ)+1/2](tanψ=

tan(x-π/4)=?题中没有说?=(tanx-tanπ/4)/(1+tanxtanπ/4)=(tanx-1)/(1+tanx)由此可以得到tanx则tan2x=2tanx/(1-tan^2x)(s

【1】f(x)=(1-sin2x)/(2cosx^2)f(x)=(1-sin2x)/(1+cos2x)1+cos2x≠0cos2x≠-12x≠π+2kπx≠π/2+kπ函数定义域:{x|x≠π/2+k

cos(π/4-x)=-4/5得(√2/2)(cosx+sinx）=-4/5所以cosx+sinx=-4√2/5-----------（1）cos(π/4+x)=sin(π/2-π/4-x)=sin(

1).f(x)=(√3)sin2x+cos2x+2=2sin(2x+π/6)+2单增区间[-π/3+kπ,π/6+kπ](2)f(C)=2sin(2C+π/6)+2=3,故C=π/3a/b=sinA/

（1）f=[根号3]/2sin2x-cos^2(x)-1/2=√3/2*sin2x-1/2(1+cos2x)-1/2=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1∵x∈[-π/

2sin2x+cos2x=1,用倍角公式展开,将1移到等号左边,化简得：2sinx(cosx-sinx)=0,cosx-sinx=0则tanx=1,x=45度,最后带进去算就行了

f(x)=根号3cos^2x+1/2sin2x=根号3/2*cos2x+1/2sin2x+根号3/2=sin(2x+π/3)+根号3/2y=sinx递减区间[2kπ+π/2,2kπ+3π/2]y=si

f(x)=(根号3/2)sin2x-cos^2x+1/4=(根号3/2)sin2x-(2cos^2x-1)/2+3/4=(根号3/2)sin2x-(cos2x)/2+3/4=sin((2x-π/6))