jcst.net
当前位置:首页 >> 求函数y=sinx%Cosx+sinxCosx的值域 >>

求函数y=sinx%Cosx+sinxCosx的值域

函数y=sinx-cosx+sinxcosx的值域为[-(1+2√2)/2,1]。 解答过程如下: y=sinx-cosx+sinxcosx =√2[(√2/2)sinx-(√2/2)cosx]+½sin2x =√2sin(x-π/4)+½cos(π/2 -2x) =√2sin(x-π/4)+½cos[2(x-π/4)] =√2sin(x-π/4)+½[1-2sin²(x...

y=sinx+cosx+sinxcosx 令sinx+cosx=T,(1) 由同角三角函数关系sinxcosx=[(sinx+cosx)^2-(sinx^2+cosx^2)]/2 把(1)式代入,得sinxcosx=(T^2-1)/2 所以y=T+(T^2-1)/2 整理得,y=1/2(T+1)^2-1 而sinx+cosx=√2sin(x+π/4)∈[-√2,√2] 所以y在T[∈-√2,√...

题目是y=sinx-cosx+sinxcosx是吧。 解: y=sinx-cosx+sinxcosx =√2sin(x-π/4)+(1/2)sin(2x) =√2sin(x-π/4)+(1/2)cos(π/2-2x) =√2sin(x-π/4)+(1/2)cos[2(x-π/4)] =√2sin(x-π/4)+(1/2)[1-2sin²(x-π/4)] =-sin²(x-π/4)+√2sin(x-π/4)+1/2 ...

如图 看样子只能用导数判断单调性 需要连续导两次,导第二次的时候可以判断导函数的单调性了 然后再判断原函数的极值

y=sinxcosx+sinx+cosx =1/2(2sinxcosx+1-1)+sinx+cosx =1/2(sinx+cosx)^2-1/2+(sinx+cosx) =1/2[(sinx+cosx)^2+2(sinx+cosx)+1-2] =1/2(sinx+cosx+1)^2-1 =1/2{(√2)*[sin(x+π/4)]+1}^2-1 所以值域为[-1,√2+1/2]

y=sinx+cosx =√2(√2/2*sinx+√2/2*cosx) =√2sin(x+π/4) ∵-1≤sin(x+π/4)≤1 ∴-√2≤y≤√2 即值域为:[-√2,√2] 行家正解,不明白可以追问!祝您学习进步 满意请点击下面的【选为满意回答】按钮,O(∩_∩)O谢谢

令t=sinx+cosx t^2=1+2sinxcosx sinxcosx=(1/2)(t^2-1) y=t+(1/2)(t^2-1) =(1/2)t^2+t-(1/2) t=sinx+cosx=√2sin(x+π/4) -√2≤t≤√2 抛物线y(t) 开口向上,对称轴为: t=-1 函数y(t)在[-√2,√2]上的单调性是先减后增,且减区间短,增区间长; 所以,...

判断y=sinx+cosx的奇偶性,非奇非偶。 当x取值为-x时,y=-sinx+cosx

y=1/sinx+1/cosx=(sinx+cosx)/sinxcosx t=sinx+cosx=√2sin(x+π/4),x在(0,π/2),t在(1,√2] 则sinxcosx=1/2*(t^2-1), y=2t/(t^2-1), 求导,y'=[2t/(t^2-1),]'列表,t在(1,√2]列区间,判断增减,再判别极值

y=(sinx+cosx)/(sinxcosx+1) 一方面: 设t=sinx+cosx=√2sin(x+π/4) ∵x属于[0,π],∴x+π/4∈[π/4,5π/4] ∴sin(x+π/4)∈[-√2/2,1] ∴t∈[-1,√2] 另一方面: (sinx+cosx)²=sin²X+cos²x+2sinxcosx=1+2sinxcosx ∴sinxcosx=(t²-1)/2 ∴y=...

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