请问为什么√2*sin(πx/8+π/4)+√2*cos(xπ/8+π/4)=2cos(πx/8).
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
函数f(x)=-√2(sin2x+π/4)+6 sin x cos x-2cos²x+1
已知sin x/2 -- 2cos x/2=0.(1)求tan x的值.(2)求 cos2x/(√2cos(π/4+x)
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x
化简sin(2π+x)cos(π-x)sin(π-x)/cos(x-π)sin(-π-x)sin(π+x)
证明下列恒等式: (1)2sin(2/π+x)cos(2/π-x)*cosθ+(2cos^2x-1)*sinθ=sin(
求化简数学公式哦[3sin^2(x/2)+cos^2(x/2)-4sin(x/2)cos(x/2)]/tan(π+x)化
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
三角函数已知f(x)=(2cos^4x—2cos²x+1/2)/(2tan(π/4—x)sin²(x
化简f(x)=(1+√2*cos(2x-π/4))/sin(π/2-x)
已知函数f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=sin(2x-π/6) ,
已知函数f(x)=(√3/2)sinπx+(1/2)cosπx,x∈R