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

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

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======以下答案可供参考======供参考答案1:sin50°(1+3tan10°)= sin50°(cos10°+3sinsin10°)/cos10°= 2sin50°sin(30°+10°)/cos10°= 2cos40°sin40/°cos10°= sin80°/cos10°=1，供参考答案2: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]=v3/4-1/2sinxmin=-1 sin50(1+√3tan10)= sin50(cos10+√3sin10)/cos10= 2sin50sin(30+10)/cos10= 2cos40sin40/cos10= sin80/cos10=1

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