Question

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？？

Answer 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°/cos10°)
=2sin40°sin50°/cos10°
=2sin40°cos40°/cos10°
=sin(2*40°)/cos10°
=sin80°/cos10°
=1

第二题
设t=x/2-π/6，则x/2+π/6=t+π/3，
原式=sin(t+π/3)cost
=(sintcosπ/3+costsinπ/3)cost
=(1/2sint+√3/2cost)cost
=1/2sintcost+√3/2cos²t
=1/4(2sintcost)+√3/4(2cos²t-1)+√3/4
=1/4sin2t+√3/4cos2t+√3/4
=1/2(sin2tcosπ/3+cos2tsinπ/3)+√3/4
=1/2sin(2t+π/3)+√3/4，易知
-1≤sin(2t+π/3)≤1，故当sin(2t+π/3)=-1时，原式取得最小值√3/4-1/2，此时
2t+π/3=-π/4+2nπ，将t=x/2-π/6代入，解之得
x=-π/4+2nπ.故
当x=-π/4+2nπ时，
sin(x/2+π/6)cos(x/2-π/6)取得最小值√3/4-1/2

Answer 2

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

Answer 3

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