求cos2x (sinxcosx)2的不定积分

y=7-4sinxcosx+4cos2x-4cos4x=7-2sin2x+4cos2x（1-cos2x）=7-2sin2x+4cos2xsin2x=7-2sin2x+sin22x=（1-sin2x）2

派再问：求过程再答：再答：给好评再问：三角形3边ABC满足B^2=AC求f(B的取值范围。。)再答：先给好评，立马帮你解决再答：应该是abc吧？再问：嗯。再问：？

sin2x=2sinxcosxcos2x=(cosx)^2-(sinx)^2sin2x+sinxcosx-cos2x=3sinxcosx-(cosx)^2+(sinx)^2由于sinx-2cosx=0

y=√3cos2x+sin2x=√[1²+(√3)²]sin(2x+z)=2sin(2x+z)其中tanz=√3/1=√3所以最大=2,最小=-2T=2π/2=π

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

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=

f(x)＝cos2x-sin2x+2√3sinxcosx＝cos2x-sin2x+√3sin2x＝cos2x+(-1+√3)sin2x＝√(5-2√3)sin(2x+φ)＝(√5-√3)si(2x+φ

f(x)=2根号3sinxcosx+cos²x-sin²xf(x)=根号2(2sinxcosx)+(cos²x-sin²x)f(x)=根号3sin2x+cos2

y=cos2x-sin2x+2sinxcosx=cos2x-2sinxcosx+2sinxcosx=cos2xx∈(0,4/π)2x∈(0,2/π)所以值域是(0,1)

x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3

f(x)=2sinxcosx＋cos2x=sin2x+cos2x=根号2*sin(2x+π/4)T=π最大值为根号2

y=sin2x+2sinxcosx-cos2x=sin2x+sin2x-cos2x=2sin2x-cos2x=(√5)sin(2x-θ）其中θ满足cosθ=2/√5,sinθ=1/√5所以周期为π,值

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

∵tanx=2，∴2sin2x-sinxcosx+cos2x=2sin2x−sinxcosx+cos2xsin2x+cos2x=2tan2x−tanx+1tan2x+1=8−2+14+1=75．

1、f(x)=-(cos²x-sin²x)+√3(2sinxcosx)=√3sin2x-cos2x=2sin(2x-π/6)T=2π/2=π递减则2kπ+π/2

f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+45°)所以最小正周期为π,最大值为√2

f(x)=sin2x+cos2x=√2sin(2x+π/4)1.f(π/4)=√2sin(2×π/4+π/4)=√2×√2/2=12.由题意,sin(a+π/4)=1/2所以a=7π/12所以sina

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=

f(x)=sin2x+cos2x=√2sin(2x+π/4)最小正周期T=2π/2=π最大值为√2再问：题目都不一样再答：哪不一样？2sinxcosx可化为sin2x呀。