sin²(π 3-x) sin²(π 6 x)

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/19 06:14:24
∫√(sin^3 x-sin^5 x)dx 上限π 下限0 求定积分

sin³x-sin^5x=sin³x(1-sin²x)=sin³xcos²x当00√(sin³xcos²x)=sinxcosx√s

化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1

2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[

已知函数f(x)=2√3sin²x-sin(2x-π/3)

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间:π/2+2

函数f(x)=sin(x+π3)sin(x+π2)

y=sin(x+π3)sin(x+π2)=(sinxcosπ3+cosxsinπ3)cosx=12sinxcosx+32cos2x=14sin2x+32•1+cos2x2=34+12sin(2x+π3

化简:(sinα-cosα)^2+sin4α/2cos2α 化简:sin^2x+sin^2(x+2π/3)+sin^2(

(sinα-cosα)²+sin4α/2cos2α=1-2sinαcosα+2sin2αcos2α/2cos2α=1-sin2α+sin2α=1sin²x+sin²(x+

sinπ(x-1)=-sinπx如何证明

sinπ(x-1)=sin(πx-π)=-sinπx

当x趋向π lim (sin 3x )/( sin 2x) 的极限为?

令a=π-x则a趋于0sin3x=sin(3π-3a)=sin3asin2x=sin(2π-2a)=-sin2a所以原式=-lim(a→0)sin3a/sin2asin3a和sin2a的等价无穷小是3

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

求∫√(sin^3x-sin^5x)dx

我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积分限是0→π∫[0→π]√(sin^3x-sin^5x)dx=∫[0→π]√[sin³x(1-sin²

已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x)

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

已知函数f(x)=2根号3sin平方x-sin(2x-π/3)

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

高中数学:已知函数f(x)=2sin(x+π/2).sin(x+7π/3)-

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3(cos*2x-sin*2x)=sin2x+根号3cos2x=2sin(2x+派

sin(x+π) sin(x+3/2π)怎么化简?

sin(x+π)=-ainx用诱导公式:x+π是第三象限的角,第三象限的角的正弦为负,π是1/2π的偶数倍,不改变函数名sin(x+3/2π)=-cosx用诱导公式:x+3/2π是第四象限的角,第四象

sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)

原式=sin(x+π/3)+√3cos(x+π/3)+2sin(x-π/3)=2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3)=2sin(x+π/3+π/6)+2

化简sin(2x+π/3)-(根号3)sin²x+sinxcosx+(根号3)/2

∵√3/2-√3sin²x+sinxcosx=(√3/2)(1-2sin²x)+(1/2)sin2x=sin(π/3)cos2x+cos(π/3)sin2x=sin(2x+π/3)

函数y=sin(x+π3

由题意x∈[0,π2],得x+π3∈[π3,5π6],∴sin(x+π3)∈[12,1]∴函数y=sin(x+π3)在区间[0,π2]的最小值为12故答案为12

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x