Question

Find the value of cos(2π/7) + cos(4π/7) + cos(6π/7) ?​

Answer 1

Step-by-step explanation:

Hello...

This is a streamlined version of the proof I posted earlier. The trig identities used are:

[a] sin (π-x) = sin x ----used in step 9

[b] sin (π+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(2π/7) + cos(4π/7) + cos(6π/7) =

2. 2sin(2π/7)[(cos(2π/7) + cos(4π/7) + cos(6π/7)] / 2sin(2π/7) =

3. [2sin(2π/7)cos(2π/7) + 2sin(2π/7)cos(4π/7) + 2sin(2π/7)cos(6π/7)] / 2sin(2π/7) =

4. [sin(4π/7) + 2sin(2π/7)cos(4π/7) + 2sin(2π/7)cos(6π/7)] / 2sin(2π/7) =

5. [sin(4π/7) + sin(6π/7) + sin(-2π/7) + sin(8π/7) + sin(-4π/7)] / 2sin(2π/7) =

6. [sin(4π/7) + sin(6π/7) - sin(2π/7) + sin(8π/7) - sin(4π/7)] / 2sin(2π/7) =

7. [sin(6π/7) - sin(2π/7) + sin(8π/7)] / 2sin(2π/7) =

8. [sin(6π/7) - sin(2π/7) - sin(π/7)] / 2sin(2π/7) =

9. [sin(π/7) - sin(2π/7) - sin(π/7)] / 2sin(2π/7) =

10. -sin(2π/7) / 2sin(2π/7) =

11. -1/2

@hardikrakholiya21

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Answer 2

Answer:

cos(2π/7) + cos(4π/7) + cos(6π/7) = -1/2.

✨✨Hope it will be helpful.✨✨