若x属于[-π 4,3π 4],求函数y=arcsin(cosx)

1、f(-π/6)=√2sin(-π/6-π/12)=√2sin(-π/4)=√2*(-√2/2)=-12、因为sinθ=-4/5,θ属于(3π/2,2π)所以cosθ=3/5f(2θ+π/3)=√2

（1）f(x）最小=(2*4*12-4^2)/2*4=10所以x属于R时,值域为【10,+∞）（2）f(x）对称轴为x=-（2*2)/4=-1又a=2>0所以x属于【1,3】时,f(x）最大=2*3^

令t=tanx因为x∈[π/4,π/3],可得：tanx∈[1,√3]即：t∈[1,√3]所以,y=-t²+10t-1,t∈[1,√3]开口向下的二次函数,对称轴为t=5所以,当t=1时,y

f(x)=2√3sinxcosx+2cosx^2-1=√3sin(2x)+cos(2x)=2sin(2x+π/6）f(x0)=2sin(2x0+π/6)=6/5sin(2x0+π/6)=3/5sin2

x属于（π/3,3π/4）∴sinx∈(√2/2,1]∴y=sin²x-sinx+1=(sinx-1/2)²+3/4令t=sinx∴y=(t-1/2)²+3/4函数在(√

答案写错了吧!f(x)=sinx–tanx求导：f′(x)=cosx-1/cos²x=(cos³x-1)/cos²≤0,定义域上单调减在区间【-π/4,π/4】最小值f(

fx=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2（α）-1fx=cos2x+根号3s

1sin(x-π/4)=√(1-cos²(x-π/4))=7√2/10∴sinx=sin[(x-π/4)+π/4]=sin(x-π/4)cosπ/4+cos(x-π/4)sinπ/4=4/5

y=cos2x+sin2x=√2(√2/2*cos2x+√2/2sin2x)=√2(sinπ/4cos2x+cosπ/4sin2x)=√2sin(π/4+2x)0

设t=cosx则y=t²+3t+2∵x属于（3π/4,π）∴t∈(-1,-√2/2)∴y=t²+3t+2=(t+3/2)²-1/4在(-1,-√2/2)上单调递增∴t=-

∵tana,tanB是方程x^2+3√3x+4=0的两根由韦达定理得：tana+tanB=-3√3,tana*tanB=4∴tana

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

cosx=-3/5,所以,sin(x-60°)=sinxcos60°-cosxsin60°=0.9196；补充问题;公式：tan（α+β）=（tanα+tanβ）／（1－tanαtanβ）带入已知条件

y=（1/cosx）+2tanx+1=secx+2tanx+1=tanx+2tanx+2=(tanx+1)+1由：x∈[-π/3,π/4]→1-√3≤tanx+1≤2故：当x=-π/4时,ymin=0

y=3sin^2x-4cosx+1y=3(1-cos^2x)-4cosx+1=3-3cos^2x-4cosx+1=-3cos^2x-4cosx+4=-3(cosx-2/3)^2+16/3.因为x属于[

令t=2x+π/4则7π/12

π/2+2kπ再问：这个x属于(0,π)和K=0有啥关系。。。。再答：不要管他有啥关系，就是π/8+kπ

x∈(π/2,π)sinx=3/5所以cosx=-√[1-(3/5)^2]=-4/5所以tanx=sinx/cosx=-3/4所以tan(x+π/4）=(-3/4+1)/[1-(-3/4)*1]=1/

f（x）=cos2x-sinx=1-2sin^2x-sinx=1-2(sin^2x+1/2sinx+1/16)+1/8=9/8-2(sinx+1/4)^2当sinx+1/4=0时,sinx=-1/4>

当x∈[-π/3,π/4]时-√3