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计算2sin50°(1+tAn60°tAn10°)结果

显然 1+√3tan10=1+tan60*tan10 =(cos60*cos10+sin60*sin10)/(cos60*cos10) =2cos(60-10)/cos10 =2cos50/cos10 于是原式 =(2sin50+2sin10*cos50/cos10) * (√2 *sin80) =2√2 *(sin50*cos10+sin10cos50) *(sin80/cos10) =2√2 *sin(50+10) =2√2 *√3/...

显然1+根号3tan10 =1+tan60tan10 =(cos60cos10+sin60sin10)/cos60cos10 =cos(60-10)/cos60cos10 =cos50/cos60cos10 =2cos50/cos10 而1+cos20=2(cos10)^2 故得到根号(1+cos20)=根号2 *cos10 原式=(2sin50+ 2sin10*cos50/cos10) * (根号2 *cos10) ...

sin50°(1+根号3tan10°) = sin50°(cos10°+根号3sin10°) / cos10° = sin50° * 2(cos60°cos10°+sin60°sin10°) / cos10° = sin50° * 2 cos(60°-10°) / cos10° = 2 sin50°cos50° / cos10° = sin100° / cos10° = sin(180°-80°) / cos10° = sin80°...

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