The value of ∑ k = 1 13 1 sin ⁡ π 4 + k - 1 π 6 sin ⁡ π 4 + k π 6 is equal to

Step-by-step solution

Let π 4 + ( K − 1 ) π 6 = θ ⇒ ∑ k = 1 13 1 sin θ sin ( θ + π 6 ) = 1 sin π 6 ∑ k = 1 13 sin π 6 sin θ sin ( θ + π 6 ) = 2 ∑ k = 1 13 sin ( θ + π 6 − θ ) sin θ sin ( θ + π 6 ) = 2 ∑ k = 1 13 sin ( θ + π 6 ) cos θ − cos ( θ + π 6 ) sin θ sin θ sin ( θ + π 6 ) = 2 ∑ k = 1 13 ( cot θ − cot ( θ + π 6 ) ) = 2 ∑ k = 1 13 ( cot ( π 4 + ( k − 1 ) π 6 ) − cot ( π 4 + k π 6 ) ) = 2 ( cot π 4 − cot ( π 4 + π 6 ) + cot ( π 4 + π 6 ) − cot ( π 4 + 2 π 6 ) + − − − − − cot ( π 4 + 12 π 6 ) − cot ( π 4 + 13 π 6 ) ) = 2 [ cot π 4 −