zxpr.net
当前位置:首页 >> 2Cos^2x >>

2Cos^2x

2cos2X=2(cosX)^2-2(sinX)^2 =4*(cosX)^2-2 =2-4*(sinX)^2

解y=2cos^2x+5sinx-4 =2(1-sin^2x)+5sinx-4 =-2sin^2x+5sinx-2 令t=sinx 则t属于[-1,1] 则y=-2t^2+5t-2 =-2(t-5/4)^2-21/8 故t=1时,y有最大值y=-2+5-2=1 t=-1时,y有最小值y=-2-5-2=-9 故函数的值域为[-9,1]

解:cos^2x-sin^2x=cos x.cos x-sin x.sin x =cos(x+x) =cos 2x

lim(x->π/3) [8(cosx)^2 -2cosx -1]/[(cosx)^2+cosx -1] =[8(1/2)^2 -2(1/2) -1]/[(1/2)^2+1/2 -1] =(2 -1 -1)/(1/4+1/2 -1) =0

见图 解:(I)f(x)==sin2x+cos2x=sin(2x+). 令 2kπ-≤(2x+)≤2kπ+,可得 kπ-≤x≤kπ+,k∈z. 即f(x)的单调递增区间为[kπ-,kπ+],k∈z. (II)在△ABC中,由,可得sin(2A+)=,∵<2A+<2π+, ∴<2A+= 或,∴A= (或A=0 舍去). ∵b,a,c成...

[-2,9/8] 解: y =2cos²x+sinx-1 =2(1-sin²x)+sinx-1 =-2sin²x+sinx+1 =-2(sinx+1/4)²+1+1/8 =-2(sinx+1/4)²+9/8 令t=sinx(-1≤t≤1) f(t)=sinx(-1≤t≤1) f(-1/4)=9/8 f(-1)=-2 f(1)=0 ∴ y=2cos^2x+sinx-1的值域是 [-2,...

第一问 第二问正

cosx=cos2(x/2)=2cos^2(x/2)-1 所以, cos^2(x/2)=[1+cosx]/2

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