关于三角函数的求证?求证:[2-2*sin(x+3*pi/4)*cos(x+pi/4)]/(cos^4-sin^4)=(

2个回答

  • 左边分母上应该是[(cosx)^4-(sinx)^4]吧!只有这样才能得证.

    首先,sin(x+3π/4)=cos(x+π/4)

    所以左边={2-2[cos(x+π/4)]^2}/{[(sinx)^2+(cosx)^2][(cosx)^2-(sinx)^2]}

    =2[sin(x+π/4)]^2/[(cosx)^2-(sinx)^2]

    =[1-cos(2x+π/2)]/cos2x

    =[1-cos(2x+π/2)]/sin(2x+π/2)

    =tan(x+π/4)(半角公式:tan(x/2)=(1-cosx)/sinx)

    =(1+tanx)/(1-tanx)

    =右边

    所以等式成立.