$$\sin^2(4) + \sin^2(8) + \sin^2(12) + ... + \sin^2(176)$$
Where the number is in degrees not radians.
$$\cos(x) = \sin(90 - x) \implies \cos(x) = \sin(90 + x)$$
$$\implies \sin(x) = \cos(x - 90)$$
$$S = \sum_{n=1}^{44} \sin^2(4n)$$
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Amad27
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The exact answer is $21.9720...$ and an approximate answer is given by $7 \cdot \pi$ – Zach466920 Jun 24 '15 at 20:27
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1You should post your work so that people interested in answering know what focus in on. – Zach466920 Jun 24 '15 at 20:27
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$$ S_1 = \sum_{n=1}^{44}\sin^2(4n) = \sum_{n=1}^{44} \frac{1-\cos (8n)}{2} =\sum_{n=1}^{44} \frac{1}{2} - \frac{1}{2}\sum_{n=1}^{44} \cos 8n = 22-\frac{1}{2}S $$ now we have $$ S = \mathcal{Re}\left[\sum_{n=1}^{44} \left(\mathrm{e}^{i8} \right)^n\right] $$ This sum is a geometric sum so we find $$ S = \mathcal{Re}\left[\frac{\mathrm{e}^{i8}-\mathrm{e}^{i8\cdot 45}}{1-\mathrm{e}^{i8}}\right] $$ the argument is $$ \frac{\mathrm{e}^{i8}-\mathrm{e}^{i8\cdot 45}}{1-\mathrm{e}^{i8}} = \frac{\left(\mathrm{e}^{i8}-\mathrm{e}^{i8\cdot 45}\right)\left(1-\mathrm{e}^{-i8}\right)}{2 -\mathrm{e}^{-i8}-\mathrm{e}^{i8}} = \frac{\mathrm{e}^{i8}-1-\mathrm{e}^{i8\cdot 45} + \mathrm{e}^{i8\cdot 44}}{2-2\cos 8} $$ thus $$ S = \mathcal{Re}\left[\frac{\mathrm{e}^{i8}-\mathrm{e}^{i8\cdot 45}}{1-\mathrm{e}^{i8}}\right] = \frac{\cos(8)-1-\cos(8\cdot 45)+\cos(8\cdot 44)}{2(1-\cos(8))} $$ putting all together we find $$ S_1 \approx 21.9720254701 $$
Chinny84
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