希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()函数f(x)=sin2x+2根号(2)cos(π/4+x) + 3 的最小值是 _________. ------------------------------------------------------------------------------------解:f(x)=sin2x+2根号(2)cos(π/4+x)
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![希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()函数f(x)=sin2x+2根号(2)cos(π/4+x) + 3 的最小值是 _________. ------------------------------------------------------------------------------------解:f(x)=sin2x+2根号(2)cos(π/4+x)](/uploads/image/z/14282969-41-9.jpg?t=%E5%B8%8C%E6%9C%9B%E6%9C%89%E4%BA%BA%E6%84%BF%E6%84%8F%E5%B8%AE%E5%BF%99%29%E5%87%BD%E6%95%B0f%28x%29%3Dsin2x%2B2%E6%A0%B9%E5%8F%B7%282%29cos%28%29%E5%87%BD%E6%95%B0f%28x%29%3Dsin2x%2B2%E6%A0%B9%E5%8F%B7%282%29cos%28%CF%80%2F4%2Bx%29+%2B+3+%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AF++_________.++------------------------------------------------------------------------------------%E8%A7%A3%3Af%28x%29%3Dsin2x%2B2%E6%A0%B9%E5%8F%B7%282%29cos%28%CF%80%2F4%2Bx%29)
希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()函数f(x)=sin2x+2根号(2)cos(π/4+x) + 3 的最小值是 _________. ------------------------------------------------------------------------------------解:f(x)=sin2x+2根号(2)cos(π/4+x)
希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()
函数f(x)=sin2x+2根号(2)cos(π/4+x) + 3 的最小值是 _________.
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解:
f(x)=sin2x+2根号(2)cos(π/4+x) + 3 = sin2x + 2(cosx - sinx) + 3.
令cosx - sinx = t, t∈[-√2,√2]
则, f(x) = g(t) = - t^2 + 2t + 4 = - (t - 2)^2 + 5 .
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请问...
这步 t∈[-√2,√2] 是怎么计算来的.
又怎么到 f(x) = g(t) = - t^2 + 2t + 4 = - (t - 2)^2 + 5 . 这步的?.
希望有人愿意帮忙,想要详细演算过程步骤..非常感谢.,短时间里一定采纳,谢谢了.
希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()函数f(x)=sin2x+2根号(2)cos(π/4+x) + 3 的最小值是 _________. ------------------------------------------------------------------------------------解:f(x)=sin2x+2根号(2)cos(π/4+x)
f(X)=sin2x+2√2cos(π/4+x)+3
=cos(2x-π/2)+2√2cos(π/4+x)+3
=1-2sin2(x-π/4)-2√2sin(x-π/4)+3
=1-[2sin2(x-π/4)+2√2sin(x-π/4)]+3
=4-[2sin2(x-π/4)+2√2sin(x-π/4)+1]+1
=5-[2sin2(x-π/4)+2√2sin(x-π/4)+1]
=5-[√2sin(x-π/4)+1]^2
-1
sin2x=2sinxcosx=-(cosx-sinx)²+sin²x+cos²x=1-t²
f(x)=g(t)=1-t²+2t+3=-t²+2t+4
t=cosx-sinx=√2 cos(x+π/4)∈[-√2,√2]
方法一:cos(x)-sin(x)=根号(2)*cos(x+π/4),-1=