f(x)=(sinx cosx)² cos²x

f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T

f(x)=sinx+(cosx/sinx)cosx;f(x)=sin(x)+(1-sinx*sinx)/sinx;f(x)=sinx+1/sinx-sinx;f(x)=1/sinx;望能看懂!

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

解析：函数有意义,sinx≠0,cosx≠0所以-1

.函数f(x)=sinxcosx=1/2sin2x,最小值=-1*1/2=-1/22.函数f(x)=sinx+sin(2x+∏/2)=sinx+cos2x=sinx+1-2sin^x=-2(sinx-

f(x)=cos^2x+(1/2)sin2x.f(x)=(1+cos2x)/2+(1/2)sin2x.=(1/2)(sin2x+cos2x)+1/2.∴f(x)=(1/2)√2sin(2x+π/4)+

=(1+cos2x)/2+sin2x/2=根号2/2sin(2x+pi/4)+1/2

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

怎么感觉cosx应该是平方啊再问：嗯的，打错了再答：(cosx)^2=(1+cos2x)/2sinxcosx=1/2*sin2x所以原式=(1+cos2x)/2-根号3/2*sin2x+1=1/2*c

（Ⅰ）由题意知，f（x）=cos4x-2sinxcosx-sin4x＝(cos2x+sin2x)(cos2x−sin2x)−sin2x＝cos2x−sin2x＝2cos(2x+π4)∴f（x）的最小正

=sinxcosx/1+sinxcosx/sin^2x+sinxcosx/cos^2x=sinxcosx/1+sinxcosx/1=sinxcosx/2=2sinxcosx/4=sin2x/4sin2

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

再答：您好，很高兴能回答您的问题，希望对您有帮助！答案见上图。很高兴为你解答，仍有不懂请追问，满意请采纳，谢谢！----【百度懂你】团队提供

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

已知函数f(x)=sinxcosx-m(sinx+cosx)(1)求函数f(x)的最小值g(x)(2)若函数f(x)在区间[π/2,π]上是单调函数,求实数m的取值范围.(1)解析：∵f(x)=sin