Prove that cos(2pi/7)+cos(4pi/7)+cos(6pi/7) = -1/2

Asked by Jerrin Thomas | 15th May, 2013, 08:57: PM

Expert Answer:

[a] sin (?-x) = sin x ----used in step 9 
[b] sin (?+x) = -sin x ----used in step 8 
[c] sin (-x) = -sin x ----used in step 6 (twice) 
[d] sin 2x = 2 sin x cos x ----used in step 4 
[e] sin x cos y = (1/2)sin(x+y) + (1/2)sin(x-y) ----used in step 5 (twice)

1. cos(2?/7) + cos(4?/7) + cos(6?/7) =
2. 2sin(2?/7)[(cos(2?/7) + cos(4?/7) + cos(6?/7)] / 2sin(2?/7) =
3. [2sin(2?/7)cos(2?/7) + 2sin(2?/7)cos(4?/7) + 2sin(2?/7)cos(6?/7)] / 2sin(2?/7) =
4. [sin(4?/7) + 2sin(2?/7)cos(4?/7) + 2sin(2?/7)cos(6?/7)] / 2sin(2?/7) =
5. [sin(4?/7) + sin(6?/7) + sin(-2?/7) + sin(8?/7) + sin(-4?/7)] / 2sin(2?/7) =
6. [sin(4?/7) + sin(6?/7) - sin(2?/7) + sin(8?/7) - sin(4?/7)] / 2sin(2?/7) =
7. [sin(6?/7) - sin(2?/7) + sin(8?/7)] / 2sin(2?/7) =
8. [sin(6?/7) - sin(2?/7) - sin(?/7)] / 2sin(2?/7) =
9. [sin(?/7) - sin(2?/7) - sin(?/7)] / 2sin(2?/7) =
10. -sin(2?/7) / 2sin(2?/7) =
11. -1/2

Answered by  | 16th May, 2013, 05:52: AM

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