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Find the value of cos(2π/7) + cos(4π/7) + cos(6π/7) ? |
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Answer» 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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