sinxcosx+cosxsinx公式
答:y =sin(x+π/3)sin(x+π/2)=-[sinxcos(π/3)+cosxsin(π/3)]cosx =-(1/2)sinxcosx-(√3/2)(cosx)^2=-(1/4)sin2x-(√3/4)(1+cos2x)=-1-(1/4)sin2x-(√3/4)cos2x=-1-(1/2)[sin2xcos(π/3)+cos2xsin(π/3)]=-...
答:于是得(1/2)sin2α+(√3/2)cos2α+2sin2α=0,即有(5/2)sin2α+(√3/2)cos2α=0,∴tan2α=-√3/5.(2).当α=π/4时,b•c=2sinxcosx+2sinαcosx+2cosαsinx=sin2x+(√2)(sinx+cosx)设y=sin2x+(√2)(sinx+cosx),再令y′=2cos2x+(√2)(cosx-sinx)=2(cos&...
答:sinx+cosx =√2(√2/2sinx+√2/2cosx) [提取√(1²+1²)=√2,使里面sinx,cosx的系数为同一个角45º的余弦和正弦]=√2(sinxcos45º+cosxsin45º) [反用sin(a+b)=sinacosb+cosasinb的公式化简】=√2sin(x+45º)一般形式:asinx+bcosx =√(...
答:第一个等于号,恒等变形 第二个等于号,cos(π/4)=sin(π/4)=√2/2
答:f(x)是个分段函数:(1)当2kπ+π/4≤x≤2kπ+5π/4(k∈Z)时,sinx≥cosx 所以此时,f(x)=sinx,值域是[-√2/2,1];(2)当2kπ-3π/4<x<2kπ+π/4(k∈Z)时,sinx<cosx 所以此时,f(x)=cosx,值域是(-√2/2,1];综合得值域是[-√2/2,1]。
答:(I)∵函数 f(x)= 3 sinx?cosx+si n 2 x = 3 2 sin2x+ 1-cos2x 2 = sin(2x- π 6 )+ 1 2 ∴函数f(x)的最小正周期为π; …(5分)由 2kπ- π 2 ≤2x- π 6 ≤2kπ+ π 2 (k∈Z)...
答:sinx+co s 2 x-si n 2 x =(2+2sinx)sinx+1-2sin 2 x=2sinx+1(14分)(2)∵f(ωx)=2sinωx+1由 2kπ- π 2 ≤ωx≤2kπ+ π 2 得 2kπ ω - π 2ω ≤x≤ 2kπ ω + π 2ω ,k∈Z ∴f...
答:奇变偶不变,符号看象限 [2] .即形如(2k+1)90°±α,则函数名称变为余名函数,正弦变余弦,余弦变正弦,正切变余切,余切变正切。形如2k×90°±α,则函数名称不变。诱导公式口诀“奇变偶不变,符号看象限”意义:k×π/2±a(k∈z)的三角函数值 (1)当k为偶数时,等于α的同名...
答:(1)y=x^0.25 y'=0.25x^(0.25-1)=0.25x^(-0.75)y'(16)=0.25/8=1/32 (2)y=sinx y'=cosx y'(π/2)=0 (3)y=cosx y'=-sinx y'(2π)=-1
答:f(x)=- 3 si n 2 x+sinxcosx = - 3 × 1-cos2x 2 + 1 2 sin2x = 1 2 sin2x+ 3 2 cos2x- 3 2 = sin(2x+ π 3 )- 3 2 ,(I) T= 2π 2 =π (II)∴ 0≤x≤ ...
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笪狠18666711691:
f(x)=sinxcos|x|+cosxsin|x| 判断函数奇偶性 -
51421陈山
: 1.f(-x) = sin(-x)cos|-x| + cos(-x)sin|-x| = -sinxcos|x| + cosxsin|x| (非奇非偶函数)2.同理可得 f(x)=lg(sinx+根号下(1+(sinx)^2) 是奇函数
笪狠18666711691:
cosx(sinx+cosx)的值域? -
51421陈山
:[答案] y=cosx(sinx+cosx) =cosxsinx+cosxcosx =(1/2)sin2x+(1+cos2x)/2 =(1/2)(sin2x+cos2x)+1/2 =(√2/2)sin(2x+π/4)+1/2 所以值域为: 【-√2/2+1/2、√2/2+1/2】.
笪狠18666711691:
sinxcosx+cosxcosx的化简 -
51421陈山
:[答案] =1/2sin2x 1/2cos2x 1/2=√2/2sin(2x л/4) 1/2
笪狠18666711691:
已知函数y=sinxcosx+sinx+cosx求值域 -
51421陈山
: ^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]
笪狠18666711691:
数学分析:sinxcosx/(sinx+cosx)的不定积分 -
51421陈山
:[答案] 方法一、 方法二、
笪狠18666711691:
sinx/cosx+cosx/sinx=4 怎么算啊 -
51421陈山
: sinx/cosx + cosx/sinx = 4 sin²x/(sinxcosx) + cos²x/(sinxcosx) = 4 (sin²x + cos²x)/(sinxcosx) = 4 1/(sinxcosx) = 4 sinxcosx = 1/4 sin2x = 1/2 2x = 2kπ + π/6 或2kπ + 5π/6 x = kπ + π/12 或 kπ + 5π/12 ,k ∈Z
笪狠18666711691:
求化简sinxcosx+(cosx)2 -
51421陈山
: sinxcosx=1/2(sin2x) cosx的平方=(1+cos2x)/2 所以=1/2(sin2x)+1/2(cosx2)+1/2
笪狠18666711691:
sinxcosx+cosxcosx的化简 -
51421陈山
: =1/2sin2x 1/2cos2x 1/2=√2/2sin(2x л/4) 1/2
笪狠18666711691:
已知函数f(x)=sinxcosx+(cosx)^2 -
51421陈山
: f(x)=sinxcosx+cos^2 x =sin2x/2+(1+cos2x)/2 =1/2*(sin2x+cos2x)+1/2 =√2/2*sin(2x+π/4)+1/2 ∴f(x)的最小正周期T=2π/2=π ∵f(x)的单调增区间为[-π/4+2kπ,3π/4+2kπ],k∈Z 当k=1时,单调增区间为[-π/4,3π/4],且[-π/6,π/3]在此区间内 ∴此时f(x)的最大...
笪狠18666711691:
sinxcosx+cosx^2 - 2 化简 -
51421陈山
: sinxcosx+cosx^2-2=(1/2)*2sinxcosx+(1/2)*[2*(cosx)^2-1]-(3/2)=(1/2)sin2x+(1/2)cos2x-(3/2)=(√2/2)sin(2x+π/4)-(3/2)