sin(2n+1)θ=sin(2n−1)θ+sinθ, then θ belongs to

sin(2n+1)θ=sin(2n−1)θ+sinθ.2sin2θcos2nθ−2sin2θcos2θ=02sin2θ(cos2nθ−cos2θ)=0 sin2θ=0 Or cos2nθ−cos2θ=0 θ=2nπ Or 2nθ=2nπ±2θ θ=2nπ or θ=n±14nπ

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({n+1}{2}) = ({n-1}{2}) + , then belongs to

Question

$sin (\frac{n + 1}{2}) θ = sin (\frac{n - 1}{2}) θ + sin θ$ , then $θ$ belongs to

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Alexander Discussed

sin (\frac{n + 1}{2}) θ = sin (\frac{n - 1}{2}) θ + sin θ

, then

θ

belongs to

11 mins ago

Discuss this question LIVE

11 mins ago

sin (\frac{n + 1}{2}) θ = sin (\frac{n - 1}{2}) θ + sin θ

2 sin \frac{θ}{2} cos \frac{n θ}{2} - 2 sin \frac{θ}{2} cos \frac{θ}{2} = 0

2 sin \frac{θ}{2} (cos \frac{n θ}{2} - cos \frac{θ}{2}) = 0

sin \frac{θ}{2} = 0

cos \frac{n θ}{2} - cos \frac{θ}{2} = 0

θ = 2 nπ

\frac{n θ}{2} = 2 nπ \pm \frac{θ}{2}

θ = 2 nπ

θ = \frac{4 nπ}{n \pm 1}

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Question Text	$sin (\frac{n + 1}{2}) θ = sin (\frac{n - 1}{2}) θ + sin θ$ , then $θ$ belongs to
Answer Type	Text solution:1
Upvotes	150