f(x)等于cos二分之x的平方-sin二分之xcos二分之x-二分之一
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f(x)=cosx/2(sinx/2+√3cosx/2)-√3/2=sinx/2cosx/2+√3cos²x/2-√3/2=1/2sinx+√3/2(1+cosx)-√3/2=sinxcos
原式=cos(π/2-x)=sinx再问:你怎么知道要提负号的??再答:直接也行再问:???我老是符号分不清。。直接怎么做的再答:cos(X减二分之派)=sinx直接得到行了,别问了,采纳吧
f(x)=(√3/2)sinx+(1/2)cosx+1=sin(x+π/6)+1单调减区间为2kπ+π/2≤x+π/6≤2kπ+3π/2化简得:2kπ+π/3≤x≤2kπ+4π/3,即单调减区间为[2
f(x)=√2cos(x/2)-(a-1)sin(x/2)f(-x)=√2cos(x/2)+(a-1)sin(x/2)f(x)=f(-x)a-1=0a=1f(x)=√2cos(x/2)T=2π/ω=2
解cos二分之x等于负三分之一且x属于π,2π即cosx/2=-1/3sinx/2=2√2/3sinx=2cosx/2sinx/2=-4√2/9cosx/2=2(cosx/4)^2-1(cosx/4)
y=1-2cos(πx/2)因为-1≤cos(πx/2)≤1所以:-2≤2cos(πx/2)≤2-2≤-2cos(πx/2)≤-2-1≤1-2cos(πx/2)≤3所以:函数y的最大值=3;最小值=-
f(x)=√3cos²0.5x+sin0.5xcos0.5x=√3/2(cosx+1)+1/2sinx=sin(60°+x)+√3/2若f(x)=3/5+√3/2,即sin(60°+x)+√
f(x)=√3(cos(x/2))^2+sin(x/2)cos(x/2)=(√3/2)(cosx+1)+(1/2)sinx=sin(π/6+x)+√3/2f(x)=3/5+√3/2sin(π/6+x)
由2的X次方小于等于16得x
f(x)=x^2+x+1=(x+1/2)^2+3/40≤x≤3/2当x=0时取最小值1当x=3/2时取最大值19/4
√3sinx/2+cosx/2=y/cosx/2=1/cosx/2则√3sinx/2cosx/2+cos²x/2=1√3/2sinx+1/2(cosx+1)=1sin(x+π/6)=1/2,
(1)由f(x)=cosx+根号3cos(x+二分之兀)化简得:f(X)=-2sin(x-π/6)要f(X)有最大值,则sin(x-π/6)=-1故:X-π/6=-π/2+2Kπ,K∈Z得出X=-π/
f(x)=根号3/2sin2x-cos^2x-1/2=根号3/2sin2x-(cos2x+1)/2-1/2=sin2xcosPai/6-sinPai/6cos2x-1=sin(2x-Pai/6)-1故
f(x)=根号3cos^x+sinxcosx-根号3/2=根号3*(1+cos2x)/2+sin2x/2-根号3/2所以f(派/8)=根号3*(1+cos派/4)/2+sin(派/4)/2-根号3/2
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=cos(3x/2+x/2)-2sinxcosx=cos2x-sin2x=√2(√2/2*cos2x
x√x=√x³
f(x)=x^(1/2)-(1/2)^x(x>=0)是增函数,f(0)=-1,f(1)=1/2,∴f(x)的零点唯一,在区间(0,1)内.
你好,马上给出答案,请稍等!再答:∵cosx=cos²(x/2)-sin²(x/2)∴y=cos(x/2)+sin(x/2)=√2sin(x/2+π/4)(x≠(2k+1)π(k∈
SINx/2-2cosx/2=0sinx/2=2cosx/2(sinx/2)/(cosx/2)=(2cosx/2)/(cosx/2)=tanx/2=2tanx=(2tanx/2)/(1-tanx/2*
f(x)=1/2*sinxcosx+√3/2*(sinx)^2=1/4*sin(2x)+√3/2*[1-cos(2x)]/2=1/4*sin(2x)-√3/4*cos(2x)+√3/4=1/2*[1/