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