4cosxcos(120degrees+x)cos(120degrees-x)=cos3x
In triangle ABC, if cosA-cosB=sinC,prove that triangle ABC is a right-angled triangle.
Choose 1 to finish is still ok, because i know they are quite difficult.
但最好做好二題
amaths ques(Trigo)可只做1題
2008-03-03 4:53 am
回答 (2)
2008-03-03 6:29 am
✔ 最佳答案
cosA ﹣cosB = - 2sin[(A+B)/2] sin[(A ﹣B)/2] ∴cosA ﹣cosB = sinC
- 2sin[(A+B)/2] sin[(A ﹣B)/2]
- 2 sin[(180 ﹣C)/2] sin[(A ﹣B)/2] = sinC
- 2 cos(C/2)sin[(A ﹣B)/2] = sinC
- 2 cos(C/2)sin[(A ﹣B)/2] = 2 sin(C/2) cos(C/2)
- sin[(A﹣B)/2] = sin(C/2)
sin[(B ﹣A)/2] = sin(C/2)
(B ﹣A)/2 = C/2
B ﹣A = C
B = A + C
∴A + B + C = 180
A + A + C + C = 180
A + C = 90
∴angle B is an right angle
∴triangle ABC is a right-angled triangle.
2008-03-03 5:03 am
4cos xcos(120°+x)cos(120°-x)
=2cos x[cos240°+cos2x]
=-cos x+2cos xcos2x
=-cos x+cos3x+cos(-x)
=cos3x-cos x+cos x
=cos3x
=2cos x[cos240°+cos2x]
=-cos x+2cos xcos2x
=-cos x+cos3x+cos(-x)
=cos3x-cos x+cos x
=cos3x
收錄日期: 2021-04-13 15:14:27
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