What is the solution of this kind of exercise $A=\sin2a+\sin4a+\ldots+\sin10a$
I have already tried to multiply with $2\cos \frac a2$ to solve this.
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Multiply both sides $\sin a$ – Someone May 09 '15 at 08:57
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Yes still can not – Peak Mang May 09 '15 at 09:23
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Can not devide it cuz it's going to be different which at the end is cosa/sina can not – Peak Mang May 09 '15 at 09:24
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Want a solution ? :) – Someone May 09 '15 at 09:52
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Here is another way to obtain it.
You may write $$ \begin{align} \sum_{k=1}^{n} \sin (2ka)&=\Im\left( \sum_{k=1}^{n} e^{2ika}\right)\\\\ &=\Im\left( e^{2ia}\frac{e^{2ina}-1}{e^{2ia}-1}\right)\\\\ &=\Im\left( e^{2ia}\frac{e^{ina}\left(e^{ina}-e^{-ina}\right)}{e^{ia}\left(e^{ia}-e^{-ia}\right)}\right)\\\\ &=\Im\left( e^{2ia}\frac{e^{ina}\left(2i\sin(na)\right)}{e^{ia}\left(2i\sin(a)\right)}\right)\\\\ &=\Im\left( e^{i(n+1)a}\frac{\sin(na)}{\sin(a)}\right)\\\\ &=\Im\left( \left(\cos ((n+1)a)+i\sin ((n+1)a)\right)\frac{\sin(na)}{\sin(a)}\right)\\\\ &=\frac{\sin(na)}{\sin(a)}\sin ((n+1)a). \end{align} $$
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Olivier Oloa
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Can u solve it as grade 11 answer? Actually this is higher that my ability – Peak Mang May 09 '15 at 08:59
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Please, can you tell me what specific point you don't understand. Thanks. – Olivier Oloa May 09 '15 at 09:01
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