设α为锐角,cos(x π/6)=﹣1/3

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（

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

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

由α为锐角，cosα=35得sinα=45，∴tanα=43-----（3分）又tan（α-β）=13，∴tanβ=tan[α-（α-β）]=tanα−tan(α−β)1+tanαtan(α−β)=4

cos(a+π/6)=3/5sin(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)=(4/

稍等下,再答：cosαcosπ/6-sinαsinπ/6=-1/3cosαcosα+sinαsinα=1两个解方程就得出来了再问：�Ҿ��Dz������˲���֪�������ʵ�TT再答：额...可

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

∵α为锐角，且cos（α+π6）=45，∴sin（α+π6）=35，∴cosα=cos[（α+π6）-π6]=cos（α+π6）cosπ6+sin（α+π6）sinπ6=45•32+35•12=43+

cos(a+π/6)=5/13那么sin(a+π/6)=12/13cos(a+π/6)=cosa*cos(π/6)-sina*sin(π/6)=5/13sin(a+π/6)=cosa*sin(π/6)

∵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/