f(x)＝sin(πx/4－π/6)－2cos²(πx/8)＋1＝sin(π/4)xcos(π/6)－cos(π/4)xsin(π/6)－cos(π/4)x＝√3/2sin(π/4)x－3/2cos(π/4)x＝√3sin[(π/4)x－(π/3)]T=(2π)/(π/4)=8在g(x)的圖像上任取一點(diǎn)(x,g(x)...求第1問的詳細(xì)步驟f(x)=(√3/2)sin(πx/4 )-(3/2)cosπx/4(sin后面展開，第二項(xiàng)降次擴(kuò)角公式) =√3sin(πx/4 - π/3)－2cos²(πx/8)＋1怎么化啊f(x)=sin(πx/4)cos(π/6)-sin(π/6)cos(πx/4)-[2cos²（πx/8) -1] =3^0.5/2sin(πx/4)-0.5cos(πx/4)-cos(πx/4) =3^0.5/2sin(πx/4)-3/2cos(πx/4) =[(3^0.5/2)²+1.5²]^0.5sin(πx/4+φ) tanφ=-(3/2)/(3^0.5/2)=-3^0.5φ=-π/3 ∴f(x)=√sin(πx/4-π/3)