1. Simplify the expression sin(π/9) [cos(π/9) + cos(π/3) + cos(5π/9) + cos(7π/9)].
2. In ∆ABC, prove that cosA cosB cosC = (1/4)(sin2A + sin2B +sin2C).
Trigonometric Functions
2010-11-14 11:07 pm
回答 (1)
2010-11-14 11:41 pm
✔ 最佳答案
1. sinπ/9(cosπ/9 + cosπ/3 + cos5π/9 + cos7π/9)= (1/2)sin2π/9 + (1/2)(sin4π/9 - sin2π/9) + (1/2)(sin2π/3 - sin4π/9) + (1/2)(sin8π/9 - sin2π/3)
= (1/2)sin8π/9
= (1/2)sin(π - 8π/9)
= (1/2)sinπ/9
2. sin2A + sin2B + sin2C
= 2sin(A+B)cos(A-B) + 2sinCcosC
= 2sin(π-C)cos(A-B) + 2sinCcos(π-A-B)
= 2sinC[cos(A-B)-cos(A+B)]
= -4sinCsinAsin(-B)
= 4sinAsinBsinC
So sinAsinBsinC = (1/4)(sin2A+sin2B+sin2C)
參考: me
收錄日期: 2021-04-29 00:04:42
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