nynw.net
当前位置:首页 >> 已知函数F(x)=sin(2x+π3)+sin(2x%π3)+2Cos2x... >>

已知函数F(x)=sin(2x+π3)+sin(2x%π3)+2Cos2x...

供参考。

亲,网友,您说的是不是下面的问题: 已知函数f x=sin(兀/2-x)sinx-根号3cos^2x,求周期、最值。 f(x)=1/2 sin2x-√3/2(cos2x+1) =sin(2x-π/3)-√3/2 T=2π/2=π。 f max=1-√3/2, f min=-1-√3/2. 送您 2015 夏祺 凉快

解:先用降幂公式把函数化为:f(x)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1 (1)最小值为-2,最小正周期为π (2)由f(C)=0知sin(2C-π/6)=1,从而可得C=π/3,再由余弦定理知:c^2=a^2+b^2-2abcosC 3=a^2+4a^2-2a*2acosπ/3,解得a=1,故b=2

(Ⅰ)当 φ= π 3 时, y= 3 sin(2x+ π 3 )-cos(2x+ π 3 ) = 3 2 sin2x + 3 2 cos2x- 1 2 cos2x+ 3 2 sin2x= 3 sin2x+cos2x= 2sin(2x+ π 6 ) ,列表: 2x+ π 6 π 6 π 2 π 3π 2 2π 13π 6 x 0 π 6 5π 12 2π 3 11π 12 π y 1 2 0 -2 0 1 故函数y=f(x...

∵f(x)=cos(2x+φ)=sin[π2+(2x+φ)]=sin(2x+π2+φ),∴f(x-π2)=sin[2(x-π2)+π2+φ)]=sin(2x-π2+φ),又f(x-π2)=sin(2x+π3),∴sin(2x-π2+φ)=sin(2x+π3),∴φ-π2=2kπ+π3,∴φ=2kπ+5π6,又-π≤φ<π,∴φ=5π6.故选:A.

f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x =根号下3sin2x/2+cos2x/2-cos2x/2+根号下3sin2x/2+2cos²x-1+1 =根号下3sin2x+cos2x+1 =2(根号下3sin2x/2+cos2x/2)+1 =2sin(2x+π/6)+1 f(π/12)=2sin(π/3)+1=根号下3+1

(1)∵f(x)=cos(2x+3π2)=sin2x,∴依题意,有g(x)=sin(x-π6),由π2+2kπ≤x-π6≤3π2+2kπ得:2π3+2kπ≤x≤5π3+2kπ,k∈Z.∴g(x)=sin(x-π6),且它的单调递减区间为[2π3+2kπ,5π3+2kπ]k∈Z.(2)由(1)知,g(A)=sin(A-π6)=13,∵0<A<π,...

(Ⅰ)f(x)=3cos2x+2sin(3π2+x)sin(π-x)=3cos2x-2cosxsinx=3cos2x-sin2x=2(32cos2x-12sin2x)=2cos(2x+π6),∴T=2π2=π,令2x+π6=kπ(k∈Z),即x=kπ2-π12(k∈Z),∴函数f(x)的对称轴方程为x=kπ2-π12(k∈Z),(Ⅱ)∵f(x)=2cos(2x+π6)...

(本小题满分14分)解:(1)f(x)=4sinx?1?cos(π2+x)2+cos2x?1=2sinx(1+sinx)-2sin2x=2sinx.∵f(ωx)=2sinωx在[?π2,2π3]是增函数,∴[?π2,2π3]?[?π2ω,π2ω]?2π3≤π2ω,∴ω∈(0,34](2)[12f(x)]2?mf(x)+m2+m?1=sin2x-2msinx+m2+m-1>0因为x∈[π...

f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4) =cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)] =cos(2x-π/3)+2sin(x-π/4)cos(π/4-x) =cos(2x-π/3)+2sin(x-π/4)cos(x-π/4) =cos(2x-π/3)+sin(2x-π/2) =cos2x*(1/2)+sin2x*(√3/2)-cos2x =(√3/2)*sin2x-(1/2)*...

网站首页 | 网站地图
All rights reserved Powered by www.nynw.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com