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tAn15度等于

tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3 亲,答题不易。记得好评奥~

tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3

tan15°=2-√3. 解:作⊿ABC,使∠C=90°,∠ABC=30°.设AC=1,则AB=2AC=2,BC=√(AB²-AC²)=√3. 延长CB到D,使BD=BA=2,连接AD. ∴∠D=∠BAD=(1/2)∠ABC=15°.(三角形外角的性质) ∴tan∠D=AC/DC,即tan15°=1/(2+√3)=2-√3.

答: tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3

求采纳

1+tan15度除以1-tan15度 =(tan45°+tan15°)/(1-tan45°tan15°) =tan(45°+15°) =tan60° =√3

tan15° =tan(60°-45°) =(tan60°- tan45°)/(1+ tan60°*tan45°) =(根号3 -1)/(1+根号3) =(根号3-1)*(根号3-1)/[(根号3+1)(根号3-1)] =(4-2根号3)/2 =2-根号3 tan60°=根号3 tan60°*tan15°=根号3*(2-根号3)=2*根号3-3

(1-tan15度)/(1+ tan15°) =(tan45°-tan15度)/(1+tan45° tan15°) =tan(45°-15°) =tan30° =根号3/3

解: tan15° =sin30°/(1+cos30°) =1/2/(1+√3/2) =2-√3 ≈0.268

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