trigonometry

👨🏻‍🏫 學術台數學

Larry

2011-05-28 11:09 pm

If A+B+C=180°,prove that cos3A+cos3B+cos3C=1-4sin(3A/2)sin(3B/2)sin(3C/2)

回答 (1)

hawk_wing_1999

2011-05-28 11:37 pm

✔ 最佳答案

RHS=1-4sin(3A/2)sin(3B/2)sin(3C/2)
=1+2[cos(3A+3B)/2-cos(3A-3B)/2]sin(3C/2)
=1+2[cos(3A+3B)/2-cos(3A-3B)/2]sin[270°-(3A+3B)/2]
=1-2[cos(3A+3B)/2-cos(3A-3B)/2]cos(3A+3B)/2
=1-2cos[(3A+3B)/2]cos[(3A+3B)/2]+2cos[(3A-3B)/2][cos(3A+3B)/2]
=1-cos(3A+3B)-cos0°+cos3A+cos3B
=cos3A+cos3B-cos(540°-3C)
=cos3A+cos3B+cos3C
=LHS

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