三角函数问题
求值:sin50°*(1+√3tan10°)
(我写得比较详细)解:sin50°*(1+√3tan10°) =sin50°*(1+tan60°tan10°) =sin50°* (tan60°-tan10°)/tan(60°-10°) =sin50°* (tan60°-tan10°)*(cos50°/sin50°) =(tan60°-tan10°)*cos50° =tan60°cos50°-tan10°cos50° =cos50°sin60°/sin60°-cos50°sin10°/cos10° 通分可得 原式=cos50°(sin60°cos10°-sin10°cos60°)/cos60°cos10° =cos50°sin(60°- 10°)/cos60°cos10° =2cos50°sin50°/2cos60°cos10° =sin100°/2cos60°cos10° =cos10°/2cos60°cos10° =1/2cos60°=1。
解:sin50°*(1+√3tan10°) =sin50°*(cos10°+√3sin10°)/cos10° =2[cos40/sin80]×[(1/2)cos10+(√3/2)sin10] =cos50/sin40=1 [(1/2)cos10+(√3/2)sin10]=cos60cos10+sin60sin10=cos50