设f(sin(x/2))=1+cosx,求f(cosx)
来源:百度知道 编辑:UC知道 时间:2024/05/17 03:19:29
解:
f[sin(x/2)]=1+cos x
=1+1-2[sin(x/2)]^2
=2-2[sin(x/2)]^2
f(cos x)=2-2(cos x)^2
解:
f(cosx)
=f[sin(x+π/2)]
=f{sin[(2x+π)/2]}
=1+cos(2x+π)
=1-cos(2x)
具体过程:
f(cosx)
=f[sin(π/2-x)]
=f{sin[(π-2x)/2]}
=1+cos(π-2x)
=1-cos(2x)
f(cosx)
=f[sin(π/2-x)]
=f{sin[(π-2x)/2]}
=1+cos(π-2x)
=1-cos(2x)
就是怎么简单、】
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