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How can one approach the evaluation of these sums? I have attempted to expand them using trigonometric identities, but I am unable to discern a viable pattern.

$$S1= \sin x \cos 2y + \sin 2x \cos 3y +... + \sin (n-1)x \cos ny$$

$$S2 = \cos x \sin 2y + \cos 2x \sin 3y +... + \cos(n-1)x \sin ny$$

Green.H
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Kiiza Bandekye
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jul 18 '23 at 06:55

1 Answers1

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I don't know if the following is what you want, but here's some compact way you can write these sums.

$$S1 + S2 = \sum_{k=2}^n (\sin(k-1)x\cos ky + \cos(k-1)x\sin ky) = \sum_{k=2}^n \sin((k-1)x + ky)$$

Similarly, $$S1 - S2 = \sum_{k=2}^n (\sin(k-1)x\cos ky - \cos(k-1)x\sin ky) = \sum_{k=2}^n \sin((k-1)x - ky)$$

Therefore,

\begin{align*} S1 & = \dfrac{1}{2} \sum_{k=2}^n \left( \sin((k-1)x + ky) + \sin((k-1)x - ky) \right)\\ S2 & = \dfrac{1}{2} \sum_{k=2}^n \left( \sin((k-1)x + ky) - \sin((k-1)x - ky) \right) \end{align*}

subroy13
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