1樓:窩巢真赤激
f(x)=2根號3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).
=(根號3)sin(x+π/2)-sin(x+π)
=(根號3)cosx + sinx 這一步用到誘導公式
=2*((根號3)/2 cosx + 1/2sinx)
=2sin(x+π/3)
將fx的影象向右平移π/6個單位
則g(x)=2sin(x+π/3-π/6)=2sin(x+π/6)
因為x+π/6在【-π/2+2kπ,π/2+2kπ】上單調遞增 在【π/2+2kπ,3π/2+2kπ】上遞減
所以x在【-2π/3 +2kπ,π/3+2kπ】上單調遞增 在【π/3+2kπ,4π/3+2kπ】上遞減
所以在【0,π】上 當x取π/3時有最大值 最大值為g(π/3)=2
當x取π時有最小值 最小值為g(π)= - 1求採納
2樓:復旦梁朝偉
f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
=√3(2sin(x/2+π/4)cos(x/2+π/4))-sin(x+π)
=√3sin(x+π/2)-(-sinx)=√3cosx+sinx
=2(sinπ/3cosx+cosπ/3sinx)=2sin(x+ π/3),
(1)2π.
(2)g(x)=2sin(x+π/6),在[0,π]上的最大值是2,最小值是-1.
已知函式f(x)已知函式f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
3樓:寒煙飛飛
解:(1)f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π) 即f(x)=√3sin(x+π/2)+sinx, f(x)=√3cosx+sinx f(x)=2(cosxsinπ/3+sinxcosπ/3) f(x)=2sin(x+π/3) 所以f(x)的最小正週期為專t=2π。 (2)因為f(x)的影象向右平移派屬/6個單位,得到函式g(x)的影象,則g(x)=2sin(x+π/3-π/6)=2sin(x+π/6), 由於π/6≦x+π/6≦7π/6 故當x=π時,g(x)最小,此時為g(π)=-1 ,當x=π/3時,g(x)最大,此時為g(π)=2
4樓:匿名使用者
f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
=√3sin(x+π/2)+sinx
=-√3cosx+sinx
=2sin(x-π/3)
(1)最小正週期t=2π/1=2π
(2)g(x)=2sin(x-π/3-π/6)=2sin(x-π/2)=-2cosx
所以g(x)在區間[0,派]上的最回大值和最答小值分別是2和-2
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