化简sin10度sin30度sin50度sin70度

化简 [cos40+sin50(1+根号3 *tan10)]/[sin70(1+cos40)] 省略了度

[cos40+sin50(1+√3 *tan10)]/[sin70(1+cos40)] = [cos40+sin50(1+√3 *sin10/cos10]/[sin70(1+cos40)] = {cos40+sin50[(cos10+√3 *sin10)/cos10]}/[sin70(1+cos40)] = {cos40+2sin50[(1/2cos10+√3/2 *sin10)/cos10]}/[sin70(1+cos40)] = {cos40+2sin50[(sin30cos10+cos30*sin10)/cos10]}/[sin70(1+cos40)] = {cos40+2sin50...全部

  [cos40+sin50(1+√3 *tan10)]/[sin70(1+cos40)] = [cos40+sin50(1+√3 *sin10/cos10]/[sin70(1+cos40)] = {cos40+sin50[(cos10+√3 *sin10)/cos10]}/[sin70(1+cos40)] = {cos40+2sin50[(1/2cos10+√3/2 *sin10)/cos10]}/[sin70(1+cos40)] = {cos40+2sin50[(sin30cos10+cos30*sin10)/cos10]}/[sin70(1+cos40)] = {cos40+2sin50[sin40)/cos10]}/[sin70(1+cos40)] = {cos40+[2cos40sin40)/cos10]}/[sin70(1+cos40)] = {cos40+[sin80)/cos10]}/[sin70(1+cos40)] = {cos40+[cos10)/cos10]}/[sin70(1+cos40)] = {cos40+1}/[sin70(1+cos40)] = 1/sin70 。
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