Jawab: cos(π/7) – cos (2π/7) + cos (3π/7) = ½ terbukti.
 

PEMBUKTIAN

Metode 1

Kita ingat identitas trigonometri:

• 2sin(α)cos(α) = sin(2α)
• 2sin(α)cos(β) = sin(α+β) + sin(α–β)
• sin(–α) = –sin(α)
• cos(π–α) = –cos(α)
• sin(π–α) = sin(α)

Misalkan x = cos(π/7) – cos (2π/7) + cos (3π/7). Maka,
x = cos(π/7) + cos(π – 2π/7) + cos(3π/7)
⇒ x = cos(π/7) + cos(3π/7) + cos(5π/7)

Kemudian, kita kalikan kedua ruas dengan 2sin(π/7).
x · 2sin(π/7) = 2sin(π/7)cos(π/7) + 2sin(π/7)cos(3π/7) + 2sin(π/7)cos(5π/7)
⇒ 2x·sin(π/7) = sin(2π/7) + sin(4π/7) + sin(–2π/7) + sin(6π/7) + sin(–4π/7)
⇒ 2x·sin(π/7) = sin(2π/7) + sin(4π/7) – sin(2π/7) + sin(6π/7) – sin(4π/7)
⇒ 2x·sin(π/7) = sin(2π/7) – sin(2π/7) + sin(4π/7) – sin(4π/7) + sin(6π/7)
⇒ 2x·sin(π/7) = sin(6π/7)
⇒ 2x·sin(π/7) = sin(π – π/7)
⇒ 2x·sin(π/7) = sin(π/7)
⇒ 2x = sin(π/7)/sin(π/7)
⇒ 2x = 1
⇒ x = ½
⇒ cos(π/7) – cos (2π/7) + cos (3π/7) = ½

∴  Terbukti bahwa cos(π/7) – cos (2π/7) + cos (3π/7) = ½.

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Metode 2

Selain identitas trigonometri di atas, kita ingat pula identitas:

• cos(2α) = cos²(α) – sin²(α)
• cos(α+β) = cos(α)cos(β) – sin(α)sin(β)

Misalkan x = cos(π/7) – cos (2π/7) + cos (3π/7).
Misalkan pula a = sin(π/7)danb = cos(π/7).

• cos(2π/7) = b²–a²
• cos(3π/7) = b(b²–a²) – a(2ab)

Maka,
x = b – (b²–a²) + b(b²–a²) – a(2ab)
⇒ x = b – b² + a² + b³ + b² – a²b – 2a²b
⇒ x = b – b² + b³ + a²(1–3b)

Karena sin²(α) = 1 – cos²(α), maka a² = 1 – b².
⇒ x = b – b² + b³ + (1–b²)(1–3b)
⇒ x = b – b² + b³ + 1 – b² – 3b + 3b³
⇒ x = 4b³ – 2b² – 2b + 1

Karena x = ½, maka:
½ = 4b³ – 2b² – 2b + 1
⇒ 1 = 8b³ – 4b² – 4b + 2
⇒ 8b³ – 4b² – 4b + 1 = 0  ....(i)

Dengan menunjukkan bahwa 8b³ – 4b² – 4b + 1 = 0, x = ½ terbukti.

Sekarang, perhatikan bahwa cos(4π/7) = –cos(3π/7).

Jika dinyatakan dalam a dan b:

• cos(4π/7) = cos²(2π/7) – sin²(2π/7)
  ⇒ cos(4π/7) = (b²–a²)² – (2ab)²
  ⇒ cos(4π/7) = (b²–1+b²)² – 4a²b²
  ⇒ cos(4π/7) = (2b²–1)² – 4(1–b²)b²
  ⇒ cos(4π/7) = 4b⁴ – 4b² + 1 – 4b² + 4b⁴
  ⇒ cos(4π/7) = 8b⁴ – 8b² + 1
• cos(3π/7) = b(b²–a²) – a(2ab)
  ⇒ cos(3π/7) = b³ – a²b – 2a²b
  ⇒ cos(3π/7) = b³ – 3a²b
  ⇒ cos(3π/7) = b³ – 3(1–b²)b
  ⇒ cos(3π/7) = b³ – 3b + 3b³
  ⇒ cos(3π/7) = 4b³ – 3b

cos(4π/7) = –cos(3π/7)
⇒ 8b^4 – 8b² + 1 = 3b – 4b³
⇒ 8b^4 + 4b³ – 8b² – 3b + 1 = 0
⇒ (b+1)(8b³ – 4b² – 4b + 1) = 0

Sudah pasti b = cos(π/7) ≠ 1, jadi 8b³ – 4b² – 4b + 1 = 0, sesuai dengan persamaan (i) di atas. Hal ini membuktikan bahwa x = ½.

∴  Terbukti bahwa cos(π/7) – cos (2π/7) + cos (3π/7) = ½.