题目

化简f(x)=根号下[（sinx/2)^4+4(cosx/2)^2]-根号下[（cosx/2)^4+4(sinx/2)^2]

提问时间：2020-11-03

答案

(sinx/2)^4=[(sinx/2)^2]^2=[1-(cosx/2)^2]^2
=1-2(cosx/2)^2+(cosx/2)^4
所以(sinx/2)^4+4(cosx/2)^2=1+2(cosx/2)^2+(cosx/2)^4
=[1+(cosx/2)^2]^2
同理(cosx/2)^4+4(sinx/2)^2=[1+(sinx/2)^2]^2
显然1+(cosx/2)^2>0,1+(sinx/2)^2>0
所以f(x)=1+(cosx/2)^2-1-(sinx/2)^2
=(cosx/2)^2-(sinx/2)^2
=cos2[2*(x/2)]
=cosx