已知f(x)=sinα-cosx,则f(α)=
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1,f(x)=sin²x+√3sinxcosx+2cos²x=1-cos²x+√3/2sin2x+2cos²x=cos²x+√3/2sin2x+1=(
你好:(1).f(x)=0.5(1-cos2x)+sin2x-0.5(1+cos2x)=sin2x-cos2x=√2sin(2x-π/4)所以最小正周期为:T=2π/2=π(2).因为函数y=sinx
f(x)=[1+cos2x)/2]/[1+(1-cos2x)/2]=(1+cos2x)/(3-cos2x)=-1+4/(3-cos2x)f'(x)=-4/(2-cos2x)^2*(2-cos2x)'=
1f(x)=√3sinπx+cosπx=2((√3/2)sinπx+(1/2)cosπx)=2sin(πx+π/3)∴最小正周期T=2π/w=2π/π=2值域f(x)∈[-2,2]2-π/2+2kπ<
f(x)=cos^4x-2sinxcosx-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x=cos2x-sin2x=根号2*cos(2x+л/4)(1)f(x)
f(x)可化为=1/2+根号3*sin(2x+pai/2)(2x+pai/2)=π/2时有最大值1/2+根号3=1/2+根号3*sin(2x+pai/3)然后(2x+pai/3)=pai/2然后sin
∵f(x)=[-sinx(-sinx)cos(π+x)]/[-2cosxsin(π-x)]=[sin²x(-cosx)]/(-2cosxsinx)=1/2sinx∴最小正周期T=2π∴函数图
f(sin(pai/2-x))=cos[3(pai/2-x)]f(cosx)=cos(3pai/2-3x)f(cospai/9)=cos(3pai/2-pai/3)=-sinpai/3=-根号3/2f
y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2
(1)f(x)=mn=(cosα+sinα)(cosα-sinα)+2√3sinαcosα=cos2α+√3sin2α=2[sin2α·cosπ/6+cos2α·sinπ/6]=2sin(2α+π/6
f(3sinx)=cos方x/sin方x=(1-sin方x)/sin方x=1/sin方x-1=9/(3sinx)方-1f(x)=9/x^2-1
f'(x)=-1/2*sin(x/2)*[sin(x/2)-cos(x/2)]+cos(x/2)[1/2cos(x/2)+1/2sin(x/2)]=-1/2*sin²(x/2)+1/2sin
f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+
f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1
你确定第一个符号是加号不是乘号?
也不知道你是f(sin(x/2))还是f((sinx)/2)?如果是第一种,利用半角公式:后面的看图,第二种不想做了.
已知f(cosx)=cos3x,则f(sinx)=要有过程sinx=cos(π/2-x)f(sinx)=f(cos(π/2-x))=cos3(π/2-x)=sin3x
解由题知函数f(x)=sinx在x∈[-π\2,π\2]是奇函数且单调递增则由f(sinα)+f(cosα-1\2)=0得f(sinα)=-f(cosα-1\2)即f(sinα)=f(-cosα+1\
f(x)=sin(2x+α)+根号3cos(2x+α)=2sin(2x+α+π/3)∵f(x)图像过(π/12,1)∴f(π/12)=2sin(π/6+π/3+α)=2sin(π/2+α)=2cosα
f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π