Question

根号下（2+2sin(2π-θ)-cos^2(π+θ))可化简为

Answer 1

解：√[2+2sin(2π-θ)-cos^2(π+θ)]
=√[2-2sinθ-cos^2θ]
=√[2-2sinθ-(1-sin^2θ)]
=√(sin^2θ-2sinθ+1)
=√(sinθ-1)^2
=1-sinθ

朋友，你刚刚采纳的那个没做完，我这个才是对的。。。。因为1＞sinθ，所以|sinθ-1|=1-sinθ
而且肯定是先用诱导公式变形后再继续运算，2sin(2π-θ)=-2sinθ，
cos^2(π+θ)=cos^2θ，这样才会越算越简单。。。好吧，晚安了。

Answer 2

=根号下（2+2sinθ-cos^2θ)
=根号下（1+2sinθ-sin^2θ)
=1-sinθ

Answer 3

楼主答的真好，学习中~~