jwbf.net
当前位置:首页 >> 化简(2Cosx四次方-2Cos²x+½)/2tAn(π/4–... >>

化简(2Cosx四次方-2Cos²x+½)/2tAn(π/4–...

=(1/2)cos(2x)

分母=2tan(π/4-x)*cos^2[π/2-(π/4+x)] =2[sin(π/4-x)/cos(π/4-x)]*cos^2(π/4-x) =2sin(π/4-x)cos(π/4-x) =sin[2(π/4-x)] =sin(π/2-2x) =cos2x 分子=1/2[4cos^4(x)-4cos^2(x)+1] =1/2(2cos^2x-1)^2 =1/2(cos2x)^2 所以原式=1/2(cos2x)^2/cos2x=1/...

[2cos^4(x)-2cos²x+1/2]/[2tan(π/4-x)·sin²(π/4+x)] =1/2*[4cos^4(x)-4cos²x+1]/{2tan[(π/2-π/4)-x]·sin²(π/4+x)} =1/2*[2cos²x-1]²]/{2tan[π/2-(π/4+x)]·sin²(π/4+x)} =1/2*[2cos²x-1]²]/{2co...

你的这个式子写得太乱了,请写正规些,这样的题应该不难! 我这里给你一些提示吧: [cos²(x)]²-cos²(x)=cos²(x)[cos²(x)-1]=﹣sin²(x)cos²(x) 其它的式子看不清。

tan(π/4-x)= (tanπ/4-tanx)/(1+tanπ/4tanx) = (1-tanx)(1+tanx) = (cosx-sinx)/(cosx+sinx) sin^2(π/4+x) = (√2/2sinx+√2/2cosx)^2= 1/2(sinx+cosx)^2 f(x) = [2cos^4x-2cos^2x+1/2]/[tan(π/4-x)sin^2(π/4+x)] =[2cos^4x-2cos^2x+1/2]/[ (...

因为(π/4-x)+(π/4+x)=π/2,所以tan(π/4-x)=cot(π/4+x) (2cos²x-1)/2tan(π/4-x)sin²(π/4+x) =sin2x/[2cot(π/4+x)sin²(π/4+x)] =sin2x/{[2cos(π/4+x)/sin(π/4+x)]sin²(π/4+x)} =sin2x/[2sin(π/4+x)cos(π/4+x)] =sin2x/sin(π/2...

sin[2π-a]cos[π+a] / (cos[a-π]cos[π/2-a] ) =-sina*(-cosa)/(-cosa*sina) =-1 (2) tan x=2,求2sin²x-sinxcosx+cos²x的值。 2sin²x-sinxcosx+cos²x =(2sin²x-sinxcosx+cos²x)/1 =(2sin²x-sinxcosx+cos²...

sin(π/4+a)=cos[π/2-(π/4+a)]=cos(π/4-a) 所以分母=2[sin(π/4-a)/cos(π/4-a)]*cos²(π/4-a) =2sin(π/4-a)cos(π/4-a) =sin[2(π/4-a)] =sin(π/2-2a) =cos2a 分子=2cos²a-1=cos2a 所以原式=cos2a/cos2a=1

(1)sin(2a-π/4)=sin2acos(π/4)-cos2asin(π/4)=√2/2(sin2a-cos2a) 原式=[1-√2*√2/2(sin2a-cos2a)]/cosa =(1-sin2a+cos2a)/cosa =(2cos²a-2sinacosa)/cosa =2cosa-2sina =2√2(√2/2cosa-√2/2sina) =2√2cos(a+π/4) (2)分母=2tan(π/4-x)*sin...

1、sin²(﹣x﹣π)cos(π+x)cosx/[tan(2π+x)cos³(﹣x-π)] =sin²x×(﹣cosx)×cosx/[tanx×(﹣cos³x)]=sin²xcos²x/(sinxcos²x)=sinx 2、tan1°×tan2°×tan3°×…×tan89° =tan1°×tan2°×tan3°×••̶...

网站首页 | 网站地图
All rights reserved Powered by www.jwbf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com