f(x)sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)的奇偶性

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f(x)sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)的奇偶性
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f(x)sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)的奇偶性
f(x)sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)的奇偶性

f(x)sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)的奇偶性
sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)=sin^4x+(1-sin^2x)^2-1/sinxcosx
=-2sin^2xcos^2x/sinxcosx=-2sinxcosx=-sin2x
所以f(x)sin^4x+cos^4x-1/sin(3π/2+x)cos(3π/2-x)为奇函数

下次问问题时,最好还是问题写清楚点比较好
我觉得函数应该是这样的吧 f(x)=(sinx)^4+(cosx)^4-1/[sin(3π/2+x)cos(3π/2-x)]=(sinx)^4+(cosx)^4-1/(cosxsinx)
f(-x)=(sin-x)^4+(cos-x)^4-1/sin(-x)cos(-x)=(sinx)^4+(cosx)^4+1/sinxcosx≠f(x)≠-f(x)
该函数不是偶函数也不是奇函数