sin+x+cos+x等于多少

阮琬姿3532已知函数y=sin²x+sinx+cosx+2,求函数y的值域 -
俟净羽13981084296 ______ 已知函数y=sin2x+sinx+cosx+2,求函数y的值域由化简sinx+cosx前分别乘以根号2*sin45.根号2*cos45.,得解根号2sinxy=sinx的平方+根好2*sinx+2令t=sinx-1=<=1y=t^2+根号2*t+2的值域即为所求,即3/2

阮琬姿3532化简sin(90度+x)+cos(270度 - x) - 2sin(派+x) - cos(90度 - x) 求解过程 -
俟净羽13981084296 ______[答案] 根据三角函数的诱导公式(必修4),把那几个公式记熟就会做这类题目了. sin(90度+x)+cos(270度-x)-2sin(派+x)-cos(90度-x) =cosx-sinx+2sinx-sinx =cosx

阮琬姿3532已知sinx+cosx= - 1/5求sin的平方x - cos的平方x的值
俟净羽13981084296 ______ 解: sin²x-cos²x=-cos2x sinx+cosx=-1/5 (sinx+cosx)²=1/25 sin²x+2sinxcosx+cos²x=1/25 1+2sinxcosx=1/25 2sinxcosx=-24/25 sin2x=-24/25 cos2x=±√(1-(24/25)²)=±√(49/625)=±7/25 sin²x-cos²x=-cos2x=±7/25 亲,请您点击【采纳答案】,您的采纳是我答题的动力,如果不明白,请追问,谢谢. 没有明白,请追问.这个算答案吗?

阮琬姿3532函数y=sin(x+π/6)+cosx的最大值是多少? -
俟净羽13981084296 ______[答案] y=sin(x+π/6)+cosx =sinxcosπ/6+cosxsinπ/6+cosx =√3/2sinx+1/2cosx+cosx =√3/2sinx+3/2cosx =√3(1/2sinx+√3/2cosx) =√3(sinxcosπ/3+cosxsinπ/3) =√3sin(x+π/3)

阮琬姿35321 - cosx=2乘以sin (x/2)的平方 -
俟净羽13981084296 ______ 这是二倍角公式1-2sin^2(x/2)=cosx 证明 cosx=cos(x/2+x/2)=cos^2(x/2)-sin^2(x/2)=1-sin^2(x/2)-sin^2(x/2)=1-2sin^2(x/2)1-cosx=2sin^2(x/2)

阮琬姿3532已知函数f(x)=sin(x+π/6)+sin(x - π/6)+cosx+a(a属于R,a是常数) -
俟净羽13981084296 ______ 1、 f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a=2sinxcosπ/6+cosx+a=√3sinx+cosx+a=√[(√3)²+1²]sin(x+z)+a=2sin(x+z)+a 其中tanz=1/√3 所以T=2π/1=2π2、 tanz=1/√3 z=π/6 f(x)=2sin(x+π/6)+a-π/2<=x<=π/2-π/3<=x+π/6<=2π/3 所以x+π/6=-π/3,f(x)最小=2sin(-π/3)+a=-√3+a x+π/6=π/2,f(x)最大=2sin(π/2)+a=2+a 所以-√3+a+2+a=√3 a=1+√3

阮琬姿3532求和Sn=cosx+cos2x+cos3x+……+cosnx -
俟净羽13981084296 ______[答案] 2sin(x/2)[cosx+cos2x+cos3x+……+cosnx ] =2sin(x/2)cosx+2sin(x/2)cos2x+2sin(x/2)cos3x+……+2sin(x/2)cosnx =sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)+……+sin(x/2+nx)-sin(nx-x/2) =sin(x/2+nx)-sin(x/2) 所以 cosx+cos2x+cos3x+……...

阮琬姿3532已知函数f(x)=sin(x+6)+sin(x - 6)+cosx+a的最大值为1.(1)常数a的值.(2)求使f(x)大于等于0成立的x 的取值集合
俟净羽13981084296 ______ f[x]=sin[x+π/6]+sin[x-π/6]+cosx+a =sinxcos[π/6]+cosxsin[π/6]+sinxcos[π/6]-cosxsin[π/6]+cosx+a =2sinxcos[π/6]+cosx+a =√3sinx+cosx+a =2sin[x+π/6]+a 当sin[x+π/6]=1,f[x]取到最大值为f[x]max=2+a=1 a=-1 )f[x]=2sin[x+π/6]-1>=0 sin[x+π/6]>=1/2 如图...

阮琬姿3532化简cos(X - 30°)+sin(X+30°)+cosX+1 -
俟净羽13981084296 ______[答案] cos(X-30°)+sin(X+30°)+cosX+1=cosX*cos30°+sinX*sin30°+sinX*cos30°+cosX*sin30°+cosX+1=[(根号3+1)/2]sinX+=[(根号3+2)/2]cosX+1

阮琬姿3532化简f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx - sinx) -
俟净羽13981084296 ______[答案] f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx-sinx) =4sinx*[1-cos(π/2+x)]/2+cos²x-sin²x =sinx(1+sinx)+cos²x-sin²x =sinx+sin²x+cos²x-sin²x =sinx+cos²x