Question

证明下列恒等式：

(1)sin4α-cos4α=2sin2α-1
(2)sinx/1-cosx=1+cosx/sinx ( /后是分母)

请写清解题过程 谢谢!

Answer 1

sin4α-cos4α=2sin(2α)*cos(2α)-(1-2*sin(2α)^2)=2sin(2α)*cos(2α)+2*sin(2α)^2-1
=2*sin(2α)(cos(2α)+sin(2α))-1
cos(2α)+sin(2α)不恒于1
sin4α-cos4α=2sin2α-1 不是恒等式

sinx/1-cosx=1+cosx/sinx 不是恒等式

追问

你错了 它们都是恒等式

追答

sin^4α-cos^4α=2sin^2α-1
左式=(sin^2α+cos^2α)(sin^2α-cos^2α)
=sin^2α-cos^2α
=sin^2α-1+sin^2α
=2sin^2α-1
=右式

sinx/(1-cosx)=(1+cosx)/sinx
左式=sinx(1+cosx)/((1+cosx)(1-cosx))
=sinx(1+cosx)/(1-cos^2x)
=sinx(1+cosx)/sin^2x
=(1+cosx)/sinx
=右式