Question

Are these trigonometric identities?

Answer 1

\[\large \sin \theta + \sin 2\theta + ...+ \sin n \theta = \frac{ \sin n(\theta/2)*\sin(n+1)(\theta/2) }{ \sin(\theta/2) }\]

Answer 2

\[\large \cos \theta + \cos 2\theta + ...+\cos n \theta = \frac{ \sin (2n+1)(\theta/2) }{ 2\sin (\theta/2) }-\frac{ 1 }{ 2 }\]

Answer 3

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Answer 4

let Sn = sin θ + sin 2θ + sin 3θ + ... + sin nθ
2 sin (θ/2) sin θ = cos (θ/2) - cos (3θ/2)
2 sin (θ/2) sin 2θ = cos (3θ/2) - cos (5θ/2)
2 sin (θ/2) sin 3θ = cos (5θ/2) - cos (7θ/2)
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2 sin (θ/2) sin nθ = cos ((2nθ - θ)/2) - cos ((2nθ + θ)/2)
thus, the sum of all terms :
(2 sin (θ/2))(Sn) = cos (θ/2) - cos ((2nθ + θ)/2)
(2 sin (θ/2))(Sn) = 2 sin ((n + 1)(θ/2)) sin ((nθ)/2)
Sn = (sin ((n + 1)(θ/2)) sin ((nθ)/2))/(sin (θ/2))
sin θ + sin 2θ + sin 3θ + ... + sin nθ = (sin ((n + 1)(θ/2)) sin((nθ)/2))/(sin (θ/2))