已知函数f(x)=根号(sin(x/2)^4+4cos(x/2)^2)-根号(cos(x/2)^4+4sin(x/2)^2)(1)化简f(x),并求f(25π/6)(2)若0

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已知函数f(x)=根号(sin(x/2)^4+4cos(x/2)^2)-根号(cos(x/2)^4+4sin(x/2)^2)(1)化简f(x),并求f(25π/6)(2)若0
xSN@~6RAO! $rib0` p "5arnMdK37|3214/Ƀ%Q3FNS"dDvv iց# THS

已知函数f(x)=根号(sin(x/2)^4+4cos(x/2)^2)-根号(cos(x/2)^4+4sin(x/2)^2)(1)化简f(x),并求f(25π/6)(2)若0
已知函数f(x)=根号(sin(x/2)^4+4cos(x/2)^2)-根号(cos(x/2)^4+4sin(x/2)^2)
(1)化简f(x),并求f(25π/6)
(2)若0<α<π,f(α)+f(α/2)=0,求α
（要过程啊~）

已知函数f(x)=根号(sin(x/2)^4+4cos(x/2)^2)-根号(cos(x/2)^4+4sin(x/2)^2)(1)化简f(x),并求f(25π/6)(2)若0
解1：
f(x)=√[sin(x/2)^4+4cos(x/2)^2]-√[cos(x/2)^4+4sin(x/2)^2]
=√{[1-cos(x/2)^2]^2+4cos(x/2)^2}-√{[1-sin(x/2)^2]^2+4sin(x/2)^2}
=√[1+2cos(x/2)^2+cos(x/2)^4]-√[1-2sin(x/2)^2+sin(x/2)^4+4sin(x/2)^2]
=√{[1+cos(x/2)^2]^2}+4-4sin(x/2)^2]-√{[1+sin(x/2)^2]^2}
=1+cos(x/2)^2-[1+sin(x/2)^2]
=1+cos(x/2)^2-1-sin(x/2)^2
=cos(x/2)^2-sin(x/2)^2
=sinx
即：f(x)=sinx
f(25π/6)=f(4π+π/6)=sin(4π+π/6)=sin(π/6)=1/2
解2：
知道了f(x),求此题就易如反掌了,就留给楼主练习吧

f(x)=√[sin(x/2)^4+4cos(x/2)^2]-√[cos(x/2)^4+4sin(x/2)^2]
=√{[1-cos(x/2)^2]^2+4cos(x/2)^2}-√{[1-sin(x/2)^2]^2+4sin(x/2)^2}
=√[1+2cos(x/2)^2+cos(x/2)^4]-√[1-2sin(x/2)^2+sin(x/2)^4+4sin(x/2)^2]

全部展开

f(x)=√[sin(x/2)^4+4cos(x/2)^2]-√[cos(x/2)^4+4sin(x/2)^2]
=√{[1-cos(x/2)^2]^2+4cos(x/2)^2}-√{[1-sin(x/2)^2]^2+4sin(x/2)^2}
=√[1+2cos(x/2)^2+cos(x/2)^4]-√[1-2sin(x/2)^2+sin(x/2)^4+4sin(x/2)^2]
=√{[1+cos(x/2)^2]^2}+4-4sin(x/2)^2]-√{[1+sin(x/2)^2]^2}
=1+cos(x/2)^2-[1+sin(x/2)^2]
=1+cos(x/2)^2-1-sin(x/2)^2
=cos(x/2)^2-sin(x/2)^2
=cosx

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