llgd.net
当前位置:首页 >> 数学:设Sin(π/4+x)=1/3,则Sin2x=? >>

数学:设Sin(π/4+x)=1/3,则Sin2x=?

sin π/2 = 1,cos 0 = 1, 局部有值(非0)直接带入就好

sin(x-∏/4) =sinx cos∏/4 - cosx sin∏/4 =sinx(1/√2) - cosx(1/√2) => sinx - cosx = -√2/3 sinx^2 + cosx^2 = 1 => sin2x = 2sinxcosx = sinx^2 + cosx^2 - (sinx - cosx)^2 = 1 - 2/9 = 7/9

cos(x-π/4)=-3/4 cos(2x-π/2)=2[cos(x-π/4)]^2-1 =1/8 sin2x=cos(2x-π/2)=1/8

展开平方即可

sin(π/4+x)=1/4 (√2/2)(sinx +cosx )= 1/4 (sinx +cosx )^2 = 1/8 1+sin2x =1/8 sin2x=-7/8

供参考。

sin(π/3 -x)=cos(x+π/6)=1/4 cos(π/3 +2x)=2cos(x+π/6)²-1=-7/8

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

解: 已知:y=sin2x 则:y'=2cos2x 令:x=π/4 有:y'=2cos2(π/4)=2cos(π/2)=0 即:切线斜率为0 故:切线与x轴平行, 又:y=sin2(π/4)=sin(π/2)=1 所以:所求切线方程为y=1

sin(π/4 - x) = 3/5sin(π/4)cosx - cos(π/4)sinx = 3/5cosx - sinx = 3√2/5cos²x - 2sinxcosx + sin²x = 18/25sin2x = 1 - 18/25 = 7/25

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