Question

If d\ne 2\pi k, (k\in Z) , prove that

\sin \theta +\sin(\theta +d)+\sin(\theta +2d)+\dots+\sin[\theta +(n-1)d] = \dfrac{\sin \dfrac{nd}{2}\sin(\theta +\dfrac{n-1}{2}d ) }{\sin \dfrac{d}{2} }

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

Using product to sum identity

\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]

\sin \dfrac{d}{2}\sin \theta = \dfrac{1}{2}[\cos( \dfrac{d}{2}-\theta )-\cos( \dfrac{d}{2}+\theta )]

\sin \dfrac{d}{2}\sin (\theta+d) = \dfrac{1}{2}[\cos( \dfrac{d}{2}+\theta )-\cos( \dfrac{3d}{2}+\theta )]

\sin \dfrac{d}{2}\sin (\theta+2d) = \dfrac{1}{2}[\cos( \dfrac{3d}{2}+\theta )-\cos( \dfrac{5d}{2}+\theta )]

\dots

\sin \dfrac{d}{2}\sin[\theta +(n-1)d] = \dfrac{1}{2}[\cos( \dfrac{2n-3}{2}d+\theta )-\cos(\dfrac{2n-1}{2}d+\theta )] )

Addition of the equations gives

\sin \dfrac{d}{2}[\sin \theta +\sin(\theta +d)+\sin(\theta +2d)+\dots+\sin[\theta +(n-1)d] ] = \dfrac{1}{2}[\cos( \dfrac{d}{2}-\theta ) -\cos(\dfrac{2n-1}{2}d+\theta )]

Using the sum to product identity

\cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

\sin \dfrac{d}{2}[\sin \theta +\sin(\theta +d)+\sin(\theta +2d)+\dots+\sin[\theta +(n-1)d] ] =\sin \dfrac{nd}{2}\sin(\theta +\dfrac{n-1}{2}d )

Therefore,

\sin \theta +\sin(\theta +d)+\sin(\theta +2d)+\dots+\sin[\theta +(n-1)d] = \dfrac{\sin \dfrac{nd}{2}\sin(\theta +\dfrac{n-1}{2}d ) }{\sin \dfrac{d}{2} }




Steven Zheng posted 1 month ago

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