三角函数问题,sin50度(1+√3tan10度)的=?sin(x/2+π/6)cos(x/2-π/6)的最小值?

sinα·cosβ=(1/2)[sin(α+β)+sin(α-β)]
sin(x/2+π/6)cos(x/2-π/6)=1/2*[sinx+sinπ/3]=v3/4-1/2
sinxmin=-1
sin50(1+√3tan10)= sin50(cos10+√3sin10)/cos10
= 2sin50sin(30+10)/cos10
= 2cos40sin40/cos10
= sin80/cos10
=1

sin50°(1+3tan10°)
= sin50°(cos10°+3sinsin10°)/cos10°
= 2sin50°sin(30°+10°)/cos10°
= 2cos40°sin40/°cos10°
= sin80°/cos10°
=1，

第一题1+√3tan10°=1+√3sin10°/cos10°=(√3sin10°+cos10°)/cos10°=2(√3/2sin10°+1/2cos10°)/cos10°=2(sin10°cos30°+cos10°sin30°)/cos10°=2sin40°/cos10°,代入原式,得原式=sin50°*(2sin40°/cos1...