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向量a=(cosx+2sinx,sinx)向量b=(cosx-sinx,2cosx) f(x)=向量a*向量b 求f(x)的单调区间
人气:137 ℃ 时间:2020-01-28 11:22:34
解答
a=(cosx+2sinx,sinx),b=(cosx-sinx,2cosx)
f(x)=a·b=(cosx+2sinx)(cosx-sinx)+2sinx*cosx
=(cosx)^2+sinxcosx-2(sinx)^2+2sinxcosx
=(cosx)^2-2(sinx)^2+3sinxcosx
=(1+cos2x)/2-2(1-cos2x)/2+(3/2)sin2x
=-1/2-(3/2)cos2x+(3/2)sin2x
=-1/2+(3√2/2)[√2/2)sin2x-(√2/2)cos2x]
=-1/2+(3√2/2)sin(2x-π/4)
函数sinx在-π/2+2kπ=所以此函数的递增区间满足-π/8+kπ=-π/8+kπ=
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