sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
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![sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2](/uploads/image/z/7018608-48-8.jpg?t=sin%5E4-cos%5E4%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E6%98%AF%E6%88%91%E9%94%99%E5%9C%A8%E5%93%AA%3Ff%28x%29%3D%28sinx%29%5E4%2B%28cosx%29%5E4%3D%5B%28sinx%29%5E2%2B%28cosx%29%5E2%5D%5E2+-2%28sinx%29%5E2%28cosx%29%5E2%3D1+-%5Bsin%282x%29%5D%5E2%2F2%3D1-%5B1-cos%284x%29%5D%2F4%3Dcos%284x%29%2F4+%2B3%2F42%CF%80%2F4%3D%CF%80%2F2%E5%87%BD%E6%95%B0%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80%2F2)
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sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
sin^4-cos^4的最小正周期是
我错在哪?f(x)=(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2
=1 -[sin(2x)]^2/2
=1-[1-cos(4x)]/4
=cos(4x)/4 +3/4
2π/4=π/2
函数的最小正周期为π/2
sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
是 f(x)=(sinx)^4+(cosx)^4
还是f(x)=(sinx)^4-(cosx)^4
如果是f(x)=(sinx)^4+(cosx)^4,你的答案没错
如果是f(x)=(sinx)^4-(cosx)^4
=[(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2]
=-cos2x
2π/2=π
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