sin(π/6+x)=1/3 cos(2π/3-2x)=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 03:38:20

sin(π/6+x)=1/3 cos(2π/3-2x)=?
sin(π/6+x)=1/3 cos(2π/3-2x)=?

sin(π/6+x)=1/3 cos(2π/3-2x)=?
cos(π/3-x)
=sin[π/2-(π/3-x)]
=sin(π/6+x)
=1/3
所以cos(2π/3-2x)
=cos[2(π/3-x)]
=2cos²(π/3-x)-1
=-7/9

已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x 化简(1)√3sin x+cos x (2)√2(sin x-cos x) (3)√2cos x-√6sin x sin(π/6+x)=1/3 cos(2π/3-2x)=? 求证:1-cos^4x-sin^4x/1-cos^6x-sin^6x=2/3 证明1-sin^6x-cos^2x/1-sin^4x-cos^4x=3/2 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin² 求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-tanX 1.已知sinα+cosα=(√2)/2,求sin^3α+cos^3α的值.(详解,注:√2表示根号2,sin^3α表示sinα的三次方)2.{cosα/(1+sinα)}-{sinα/(1+cosα)}={2(cosα-sinα)/(1+sinα+cosα)}(详解)3.已知sinα是方程5x^2-7x-6=0的根.求{cos(2 化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1 已知函数f(x)=[2sin(x-π/6)+√3sin x]cos x+sin^2x,x∈R 1-(sin^6x+cos^6x)=3sin^2xcos^2x 如何证明? 已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2 三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x 已知sin(x+π/3)=1/4,求sin(2π/3-x)+cos²(π/6-x) 为什么cos(2x-π/3)-sin(5π/6) =cos(2x-π/3)? 若sin(π+x)+cos(π/2+x)=-m,则cos(3π/2-π)+2sin(6π-x)=? 求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos^4x=sin^2x 化简sin(x+7π/4)+cos(x-3π/4)步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2-