设X为锐角,若sin(X π 6) 3 5,求

解决方案,cosxcos30°-sinxsin30°+的sinx=0/5√3/2cosx1/2sinx=0/5COS（X-30°）=8/10COS（X-60°）=COS2（X-30°）=2cos^2（

cos(X-cosX)=sin(π/2-X+cosX)带入原式sin(x+sinx)=sin(π/2-x+cosx)x+sinx=π/2-x+cosxsinx-cosx=π/2-2x对于原式左侧有si

（1）f(x)=sin²x-sin（2x-π/6)=(1-cos2x)/2-(sin2xcosπ/6-cos2xsinπ/6)=1/2-(cos2x)/2-sin2xcosπ/6+(cos2

sinα=1/2,α为锐角∴cosα=√(1-sin²α)=√(1-1/4)=√3/2∴f(α)=sin(α+π/6)=sinα*cos(π/6)+cosα*sin(π/6)=(1/2)*(

a是锐角π/2

设b=a+π/6,sinb=3/5,sin2b=2sinbcosb=24/25,cos2b=7/25sin（2a+π/12）=sin（2a+π/3-π/4）=sin（2b-π/4）=sin2bcosπ

50分之17倍根号2

a为锐角,cos(a+π/6)=3/5则：sin(a+π/6)=4/5sin(a-π/12)=sin[(a+π/6)-π/4]=sin(a+π/6)cos(π/4)-cos(a+π/6)sin(π/4

sin（2a+π/12）=sin【2(a+π/6)—π/4】=sin2(a+π/6)cos(π/4）—cos2(a+π/6)sin(π/4）因为cos(a+π/6)=4/5,a为锐角,所以0

设A+π/6=α,则2A+π/12=2α-π/4cosα=4/5,α为锐角,sinα=3/5,所以sin2α=2sinαcosα=24/25,cos2α=2(cosα)^2-1=7/25sin(2A+

设α为锐角,若cos(α+π/6)=4/5,求sin（2α+π/12）,cos(α+π/6)=4/5sin(α+π/6)=3/5,sin（2α+π/3）=24/25cos（2α+π/3）=7/25si

设x为锐角,cos(x+π除以6)=五分之四=4/5,∵4/5>√2/2,cos(π除以4)=√2/2∴0

f(x)=sin(2x+π/6)+2msinxcosx=(√3sin2x)/2+(cos2x)/2+msin2x=(m+√3/2)sin2x+(cos2x)/2所以f(x)的最大值为√[1/4+(m+

g(x)=f(x+a)=sin[2(x+a)+π/6]+3/2=sin(2x+2a+π/6)+3/2为偶函数,所以2a+π/6=π/2+kπ,k∈Za=π/6+kπ/2,k∈Z

f(x)=0.5-0.5cos2x-(√3/2*sin2x-0.5cos2x)=1/2[1-√3sin2x]sin2x=1,fmin=(1-√3)/2sin2x=-1,fmax=(1+√3)/2值域为

解函数f(x)=sin(2x+π/6)+m,函数f(x)的最小值为2即-1+m=2即m=3函数f(x)=sin(2x+π/6)+31若x∈[-π/6,π/3]时当x=1/6π时y有最大值f(x)=si

sinxcosπ/6-cosxsinπ/6=1/3√3/2*sinx=1/3+1/2*cosx平方3/4*sin²x=(1/3+1/2*cosx)²3/4(1-cos²x

∵sin(πcosx)=cos(πsinx)=sin(π/2-πsinx)∴πcosx=π/2-πsinx或πcosx+(π/2-πsinx)=π即cosx+sinx=1/2或cosx-sinx=1/

∵x∈[0，π2]，∴(2x+π6)∈[π6，7π6]．∵关于x的方程sin（2x+π6）=k+12在[0，π2]内有两个不同根α，β，∴12=sinπ6≤k+12＜1，解得0≤k＜1，∴α+β＝2×