cos(π 4 X)

y＝2cos(x+π4)cos(x−π4)+3sin2x=2(12cos2x−12sin2x)+3sin2x=cos2x+3sin2x=2sin(2x+π6)∴函数y＝2cos(x+π4)cos(x−

∵f(x)＝cos(x+π4)cos(x−π4)=12cos2x，∴其周期T=π，又图象上相邻两个对称中心的距离是T2，故答案为：π2．

-cosx

原式=[2sinxcosx＋2sin²x]/[(cosx－sinx)/cosx]=[2sinxcosx(sinx＋cosx)]/[cosx－sinx]=[sin2x][(sinx＋cosx)

sin(π/4-x)=sin(π/4)cosx-cos(π/4)sinx=（√2/2）cosx-（√2/2）sinx;cos(x+π/4)=cosxcos(π/4)-sinxsin(π/4)=（√2/

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1=-√2(

不知道,sorry

原式=（cosxcosπ/4+sinxsinπ/4）^2-（cosxcosπ/4-sinxsinπ/4）^2=1/2(1+sin2x)-1/2(1-sin2x)=sin2x如果x没有范围的话,那么原式

化简f(x)=cosx(asinx-cosx)+cos（π/2-x）cos(π/2-x).=acosxsinx-cos^2x+sin^2x=a/2sin2x-cos2x2.化简f(x)=4cosxsi

f(x)=4cos(wx-π/6)sinwx-cos(2wx+π)=(4coswxcosπ/6+4sinwxsinπ/6)sinwx+cos2wx=2√3coswxsinwx+2(sinwx)^2+1

∫(0~π)根号(cos^2x-cos^4x)dx=2∫(0~π/2)根号(cos^2x(1-cos^2x))dx=2∫(0~π/2)cosxsinxdx=2∫(0~π/2)sinxdsinx=(si

[（sinx-cosx）/(sinx+cosx)]^4=[(1-tgx)/(1+tgx)]^4=[tg(x-45)}^4=[sec^2(x-45)-1]^2由此再求,上面两答案都不对

图片正在上可能 要上些时间,现在上传图片不知道什么原因经常传不上去.

f(x)=cos^2x/sin(π/4+x)cos(x+π/4),f(5π/12)=cos²(5π/6)/[(sin2π/3)cos(2π/3)]=(3/4)/(-√3/2*1/2)=-√3

∫(π/2,-π/2)√(cos^2x-cos^4x)dx=∫(π/2,-π/2)√[cos^2x(1-cos^2x)]dx=∫(π/2,-π/2)√[cos^2x*sin^2x]dx=∫(π/2,-

Sin(x+π/4)*cos[π/4]+Cos[x+π/4]*sin[π/4]=sin[(x+π/4)+π/4]=sin(x+π/2)正弦函数的和角公式再问：正弦函数的和角公式..我们还没学到..你q

设F（X）=g(x)+h(x)+1/4g（x)=cos(π/3+x)cos(π/3-x)=[cosπ/3cosx－sinπ/3sinx][cosπ/3cosx+sinπ/3sinx]=[1/2cosx

f(x)=2√3sinxcosx-2cos(x+π/4)cos(x-π/4)=√3sin2x+2sin(x+π/4-π/2)cos(x-π/4)=√3sin2x+sin(2x-π/2)=√3sin2x

1求值：sin（π／4+3x）cos（π／3-3x）+cos（π／6+3x）cos（π／4+3x）解∵（π／3-3x）=π／2-（π／6+3x）∴原式=sin（π／4+3x）sin（π／6+3x）+c

原式=(-sinx*sinx)/(cos(1.5π-x)*sin(4.5π+x))=sinx/cosx=tanx=-3/4