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](/uploads/image/z/1729778-50-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E6%A0%B9%E5%8F%B7%28sin%28x%2F2%29%5E4%2B4cos%28x%2F2%29%5E2%29-%E6%A0%B9%E5%8F%B7%28cos%28x%2F2%29%5E4%2B4sin%28x%2F2%29%5E2%29%281%29%E5%8C%96%E7%AE%80f%28x%29%2C%E5%B9%B6%E6%B1%82f%2825%CF%80%2F6%29%282%29%E8%8B%A50)
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]
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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
收起