Question

Prove that :

sin3π/5 + sin4π/5 + sin6π/5 + sin7π/5 = 0.

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Answer 1

Answer:

Step-by-step explanation:

If A + B = π, then

⇒ A = π - B ⇒ sin A = sin(π-B) = sin A = sin B

∴ π/5 + 4π/5 = π ⇒ sin π/5 = sin 4π/5 and 2π/5 + 3π/5 = π ⇒ sin 2π/5 = sin 3π/5

∴ L.H.S. = sin π/5 sin 2π/5 sin 3π/5 sin 4π/5

= L.H.S. = sin π/5 sin 2π/5 sin 2π/5 sin π/5

⇒ L.H.S. = (sin π/5 sin 2π/5)² = (sin 36° sin 72°)² = (sin 36° sin 18°)²

⇒ L.H.S.= {(√10-2√5/4) × (√10+2√5/4)}² = (10-2√5/16) × 10+2√5/16

⇒ L.H.S = 100-20/256 = 80/256 = 5/16 = R.H.S.

L.H.S. = R.H.S. hence proved

Answer 2

Answer:

Step-by-step explanation:

For solution :

See the attachment.

Hence, it is proved.