Summation of ∑r=1ncos⁡(2rx)sin⁡((2r+1)xsin⁡((2r−1)x}My Try:S=∑r=1ncos⁡(2rx)sin⁡((2r+1)x−(2r−1)x)sin⁡2x:sin⁡((2r+1)xsin⁡((2r−1)x}S=∑r=1ncos⁡(2rx)(sin⁡((2r+1)xcos⁡(2r−1)x−cos⁡(2r−1)x)sin⁡(2r+1)x)sin⁡2xsin⁡((2r+1)xsin⁡((2r−1)x)}S=∑r=1ncos⁡(2rx)sin⁡2x(cot⁡(2r−1)x−cot⁡(2r+1)x)

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sa3b4or9i9 · Accepted Answer

Here is one possible approach. First we rewrite the numerator ascos⁡(2rx)=cos⁡[(r+12)x+(r−12)x]=cos⁡2r+12xcos⁡2r−12x−sin⁡2r+12xsin⁡2r−12xand the denominator as4sin⁡2r+12xcos⁡2r+12x⋅sin⁡2r−12xcos⁡2r−12xTherefore the original summation can be written asSn=14∑r=1n1sin⁡2r+12xsin⁡2r−12x−14∑r=1n1cos⁡2r+12xcos⁡2r−12xNow we can use the trick that multiply a factor sin⁡xsin⁡x and usesin⁡x=sin⁡2r+12xcos⁡2r−12x−sin⁡2r−12xcos⁡2r+12xFor the first sum,sin⁡xsin⁡2r+12xsin⁡2r−12x=cot⁡2r−12x−cot⁡2r+12xFor the second sum,sin⁡xcos⁡2r+12xcos⁡2r−12x=tan⁡2r+12x−tan⁡2r−12xThus the original sum isSn=14sin⁡x(cot⁡12x−cot⁡2n+12x−tan⁡2n+12x+tan⁡12x)

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