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tAn15°等于多少怎么算的?

2-根号3 在一个有15度角的直角三角形中,它的对边除以邻边(直角边)的值

解法一:cos30 °= √¯3/2 cos15 °= √¯[(1 + cos30° )/2] sin15° = √¯[(1 - cos30° )/2] tan15° = sin15° /cos15° = √¯[(1 - cos30° )/(1 + cos30° )] = √¯[(2 - √¯3)/(2 + √¯3)] = (2 - √¯3) 解法二:作...

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

解法一:cos30 °= √¯3/2 cos15 °= √¯[(1 + cos30° )/2] sin15° = √¯[(1 - cos30° )/2] tan15° = sin15° /cos15° = √¯[(1 - cos30° )/(1 + cos30° )] = √¯[(2 - √¯3)/(2 + √¯3)] = (2 - √¯3) 解法二:作...

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求采纳

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~

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