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sin60°+Cos60°+tAn60°怎么算

解: cos60°tan60° =cos60°×sin60°/cos60° =sin60° =√3/2 若是单独求cos60°、tan60°,可利用直角三角形,60°的余角为30°,30°角所对的直角边等于斜边的一半,设30°角所对的直角边为1,则斜边为2,根据勾股定理,60°角所对的直角边为√3, cos60°=...

解: cos60°tan60° =cos60°×sin60°/cos60° =sin60° =√3/2 若是单独求cos60°、tan60°,可利用直角三角形,60°的余角为30°,30°角所对的直角边等于斜边的一半,设30°角所对的直角边为1,则斜边为2,根据勾股定理,60°角所对的直角边为√3, cos60°=...

sin30°=1/2, sin45° = √2/2 , sin60°=√3/2 cos30°=√3/2, cos45° = √2/2 , cos60°=1/2 tan30°=√3/3, tan45° = 1 , tan60°=√3

sin60°=32,cos60°=12,tan60°=3,则12<32<3,即cos60°<sin60°<tan60°.故答案为:cos60°<sin60°<tan60°.

原式=(32)2-12×3+32,=34-32+32,=34.

把tan60°= 3 ,sin60°= 3 2 ,cot30°= 3 ,cos30°= 3 2 ,代入原式得:原式=(1+tan60°-sin60°)(1-cot30°+cos30°)=(1+ 3 - 3 2 )(1- 3 + 3 2 )=(1+ 3 2 )(1- 3 2 )=1-( 3 2 ) 2 =1- 3 4 = 1 4 .

原式=4-1+1?1232?3=4-1+1=4.

怎么算? sin60°=√3 /2 tan45°=1 结果不就是3√/2 -1吗

sin60等于二分之根号三,cos60等于二分之一

原式=(√3/2)/(1+√3/2)-1/(1-√3) =√3/(2+√3)+1/(√3-1) 分母有理化 =√3(2-√3)/(4-3)+(√3+1)/(3-1) =2√3-3+(√3+1)/2 =(5√3-5)/2

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