2sin(π 4 x)的平方怎样化简?

sin(2x)=2tanx/(1+tan²x)=3/5sin²(x+π/4)=[1-cos(2x+π/2)]/2=(1/2)+sin(2x)/2=(1/2)+(1/2)(3/5)=

因为sin^2x=（1-cos2x）/2所以f(X)=sin(2x-π/4)-2根号2sin^2x=√2/2sin2x-√2/2cos2x-√2(1-cos2x)=√2/2sin2x+√2/2cos2

函数f(x)=2sin^2(x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x=2sin(2x-π/3)1.当x属于R时,函数f(x)的最小正周期T

右移π/2个单位

y=sin²2xy′=2×(cos2x)×2sin2x=4cos2x·sin2x=2sin4xy〃=2(4x)′cos4x=8cos4x

因为sin²a+cos²a=1所以cos²a=1-sin²a所以左边=(sin²a+cos²a)(sin²a-cos²a

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

y=sin^x+2sinxcosx=1/2-cos2x/2+sin2x=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2设cosa=2/根号5,sina=-1/根号5上式=根号下

前提掌握：sinx*sinx+cosx*cosx=1cos2x=2*cosx*cosx-1=1-2*sinx*sinxcos（x-π/4）=-sin（x-π/4+π/2）=-sin(x+π/4)sin

sinx是周期函数,所以(sinx)^2当然也是.不过如果要基本周期应该用半角公式降次一下:(sinx)^2=(1-cos2x)/2.由cos2x的周期性可知,基本周期是π.

设f(x)=lg(sinx+根号1+sinx平方）=lga则f(-x)=lg(1/a)=lg(a^-1)=-lga=-f(x)所以为奇函数（sinx+根号1+sinx平方）*[sin（-x）+根号1+

y=sin²x+√3sinxcosx=(1-cos2x)/2+√3/2sin2x=√3/2sin2x-1/2cos2x+1/2=cosπ/6sin2x-sinπ/6cos2x+1/2=sin

(x+π/6)+(5π/6-x)=π,(x+π/6)+(π/3-x)=π/2根据诱导公式：sin(5π/6-x)=sin[π-(x+π/6)]=sin(x+π/6)=1/4sin^2(π/3-x)=s

sin^2x=1-cos^2x

利用三角函数积化和差及和差化积公式,f(x)=cos(2x-π/3）-[cos(x-π/4+x+π/2)-cos(x-π/4-x-π/2)]=cos(2x-π/3）-[cos(2x+π/4)-cos(

y=(sinx)^2-cosx=1-(cosx)^2-cosx=-(cosx+1/2)^2+5/4x∈(-π/4,π/4)cosx∈(√2/2,1)令cosx=1,得y=-1令cosx=√2/2,得y

3-4sin平方x＞04sin平方x＜3|sinx|＜√3/2-√3/2＜sinx＜√3/22kπ-π/3＜x＜2kπ+π/3k属于Z

实数范围内无法分解,在复数范围内可以.4x²+2x+1=(2x+1/2)²+3/4=(2x+1/2)²-(i√3/2)²【i²=-1】=[2x+(1+

因式分解为(sinX-1/2)^2+11/4你把原题放出来,那就简单了,sinx的定义域是(-1,1)之间,所以(sinX-1/2)^2的最大值是9/4,最小值是0,所以Y的最大值是9/4+11/4=