设x为锐角,若cos(x 6)=4 5,则sin(2x 12)

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

α∈(0,π/2)α+π/6∈(π/6,2π/3)cos(α+π/6)=4/5>0∴α+π/6∈(π/6,π/2)∴2α+π/3∈(π/3,π)cos(2α+π/3)=2cos²(α+π/6

sin(2α+π/12)=sin(α+π/6+α-π/12)=sin(α+π/6)cos(α-π/12)+cos(α+π/6)sin(α-π/12)cos(α+π/6)=4/5,α为锐角,则sin(α

sinα=3cosα则tana=3则sinαcosα=sinαcosα/(sin²a+cos²a)=tana/(tan²a+1)=3/(9+1)=3/10

a是锐角π/2

a为锐角,cos(a+π/6)=4/5所以,sin(a+π/6)=3/5sin(2a+π/3)=2sin(a+π/6)cos(a+π/6)=2×(3/5)×(4/5)=24/25cos(2a+π/3)

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

cos(2a+π/3)=2cos²(a+π/6)-1=7/25a为锐角,则：2a+π/3∈(π/3,π/2)∴sin(2a+π/3)=24/25sin(2a+π/12)=sin[(2a+π/

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

a//b,可得：a=tb,所以有：3=√3t解得：t=√3sina=cosat所以：sina/cosa=t即：tana=√3　得：a=60°

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

∵cosα+cos2α=1,∴cosα+cos²α-sin²α=cosα+cos²α-1+cos²α=1解得cosα=（-1±√17）/4,又∵α为锐角,∴co

/>（cosα-sinα）/cosα+sinα）=（1-tanα)/(1+tantanα)=(1-3)/(1+3)=-1/2

∵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/