a) If A+B+C=180°, prove that cos 3A + cos 3B + cos 3C = 1 - 4 sin 3A/2 sin 3B/2 sin 3C/2
b) Given that cos 3A + cos 3B + cos 3C = 1, where A, B and C are the interior angles of triangle ABC, show that one of the angles of the triangle is 120°.
m2 trigo
2012-05-10 2:02 am
回答 (1)
2012-05-10 2:48 am
✔ 最佳答案
a)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)
b)
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°
收錄日期: 2021-04-21 22:25:01
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