trigonometry(3)
2011-05-29 11:44 pm
It is giventhat A+B+C=180°, I show that cos3A+cos3B+cos3C=1-4sin(3A/2)sin(3B/2)sin(3C/2)Giventhat cos3A+cos3B+cos3C=1, where A,B and C are the interior angles of ΔABC,show that one of the angles of the triangle is 120°
回答 (1)
2011-05-30 12:26 am
✔ 最佳答案
cos3A + cos3B + cos3C= 2 cos(3(A+B)/2) cos(3(A-B)/2) + cos3C= 2 cos(3(180° - C)/2) cos(3(A-B)/2) + cos3C= 2 cos(270° - 3C/2) cos(3(A-B)/2) + cos3C= - 2 sin(3C/2) cos(3(A-B)/2) + 1 - 2sin²(3C/2) = 1 - 2sin(3C/2) ( cos(3(A-B)/2) + sin(3C/2) )= 1 - 2sin(3C/2) ( cos(3(A-B)/2) + sin ( 3(180° - (A+B))/2 ) )= 1 - 2sin(3C/2) ( cos(3(A-B)/2) + sin (270° - 3(A+B)/2) )= 1 - 2sin(3C/2) ( cos(3(A-B)/2) - cos(3(A+B)/2) )= 1 - 2sin(3C/2) ( - 2 sin(3A/2) sin(- 2B/2) )= 1 - 2sin(3C/2) ( 2 sin(3A/2) sin(2B/2) )= 1 - 4 sin(3A/2) sin(3B/2) sin(3C/2);cos3A + cos3B + cos3C = 11 - 4 sin(3A/2) sin(3B/2) sin(3C/2) = 1sin(3A/2) sin(3B/2) sin(3C/2) = 0sin(3A/2) = 0 or sin(3B/2) = 0 or sin(3C/2) = 0Since 0 < 3A/2 < 3*180/2 = 270° ,==>
3A/2 = 180°A = 120°Similarly , or B = 120°or C = 120°
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