(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π

问题描述:

(1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π

分子(1+sinα+cosα)(sinα/2-cosα/2)
=[2(cosα/2)^2+2sinα/2*cosα/2](sinα/2-cosα/2)
=2cosα/2(sinα/2+cosα/2)(sinα/2-cosα/2)
=2cosα/2[(sinα/2)^2-(cosα/2)^2]
=2cosα/2*(-cosα)
分母√2+2cosα
=√2+2[ 2(cosα/2)^2-1]
=√4(cosα/2)^2
=2cosα/2
所以原式=-cosα