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

Question

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

Answer 1

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.