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已知函数F(x)=sin(2x+π/6),其中x∈[%π/6,π/3)若F(x)...

你是求f(x)的值域吗?

(1) f(x)=sin(2(x-π/12)) x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6] 单调递减区间是2(x-π/12)∈[π/2,3π/2] 即x-π/12∈[π/4,3π/4] 则x∈[π/3,5π/6] (2)x∈[-π/12,π/2]时, 2(x-π/12)∈[-π/3,5π/6] 而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),...

因为x∈[-π/6,a],所以2x+π/6∈[-π/6,2a+π/6] 因为值域是[-1/2,1],画一个单位圆可知定义域的长度是小于2π的。 然后通过单位圆可知2a+π/6小于等于7π/6 ,大于等于π/2 所以a∈[π/6,π/2]

题目本身就是错的 sinA正弦函数的取值范围(-1,1)

f(x)=sin(x+π/6) x∈【-π/3,α】 x+π/6∈【-π/6,α+π/6】 f(x)的值域是【-1/2,1】 x+π/6=-π/6或x=7π/6时f(x)有最小值-1/2 x+π/6=π/2时(fx)有最大值1 ∴π/2≤α+π/6≤7π/6 ∴π/3≤α≤π 即,α的取值范围是【π/3,π】

最小正周期2π/2=π 单调减区间 2kπ+π/2≤2x+π/6≤2kπ+3π/2 2kπ+π/3≤2x≤2kπ+4π/3 kπ+π/6≤x≤kπ+2π/3 y=sin(2x+π/6)+3/2 =sin[2(x+π/12)]+3/2 因此将sin2x向左移π/12得到sin[2(x+π/12)] 再向上移3/2个单位即可

∵函数f(x)在区间π/6到2π/3上函数值从1减小到-1 ∴T/2=2π/3-π/6=π/2,∴T=π 由T=2π/w=π==>w=2 ∵x=π/6时,f(x)取得最大值1 ∴sin(2*π/6+φ)=1 ∴2*π/6+φ=kπ+π/2,k∈Z ∵|φ|

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

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