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Question Number 30743 by abdo imad last updated on 25/Feb/18

$$decomposeinsideR\left[x\right]p\left(x\right)=x^{2n+1}-1thenfind$$$$\prod _{k=1}^{n}sin\left(\frac{k\pi }{2n+1}\right).$$

Commented by abdo imad last updated on 02/Mar/18

$$therootsofp\left(x\right)?z^{2n+1}-1=0\iff z^{2n+1}=e^{i\left(2k\pi \right)}$$$$sotherootsare{z}_k=e^{i\frac{2k\pi }{2n+1}}withk\in \left[[0,2n]\right]\Rightarrow$$$$p\left(x\right)=\prod _{k=0}^{2n}(x-z_k)=(x-1)\prod _{k=1}^{2n}(x-e^{i\frac{2k\pi }{2n+1}})\Rightarrow$$$$\frac{p\left(x\right)}{x-1}=\prod _{k=1}^{2n}(x-e^{i\frac{2k\pi }{2n+1}})\Rightarrow {lim}_{x\to 1}\frac{p\left(x\right)}{x-1}=\prod _{k=1}^{2n}(1-e^{i\frac{2k\pi }{2n+1}})$$$$\Rightarrow (2n+1)=\prod _{k=1}^{2n}(1-cos\left(\frac{2k\pi }{2n+1}\right)-isin\left(\frac{2k\pi }{2n+1}\right))$$$$=\prod _{k=1}^{2n}(2{sin}^2\left(\frac{k\pi }{2n+1}\right)-2isin\left(\frac{k\pi }{2n+1}\right)cos\left(\frac{k\pi }{2n+1}\right))$$$$=2^{2n}{(-1)}^n\prod _{k=1}^{2n}sin\left(\frac{k\pi }{2n+1}\right)e^{i\left(\frac{k\pi }{2n+1}\right)}$$$$=2^{2n}{(-i)}^n\left(\prod _{k=1}^{2n}sin\left(\frac{k\pi }{2n+1}\right)\right)e^{i\frac{\pi }{2n+1}\frac{2n(2n+1)}{2}}$$$$=2^{2n}\prod _{k=1}^{2n}sin\left(\frac{k\pi }{2n+1}\right)but$$$$\prod _{k=1}^{2n}sin\left(\frac{k\pi }{2n+1}\right)=\prod _{k=1}^{n}sin\left(\frac{k\pi }{2n+1}\right)\prod _{k=n+1}^{2n}sin\left(\frac{k\pi }{2n+1}\right)the$$$$ch.k-n=jgive$$$$\prod _{k=n+1}^{2n}sin\left(\frac{k\pi }{2n+1}\right)=\prod _{j=1}^{n}sin\left(\frac{(n+j)\pi }{2n+1}\right)$$$$=\prod _{j=1}^{n}sin(\pi -\frac{n\pi +j\pi }{2n+1})=\prod _{j=1}^{n}sin\left(\frac{2n\pi +\pi -n\pi -j\pi }{2n+1}\right)$$$$=\prod _{j=1}^{n}sin\left(\frac{(n+1-j)\pi }{2n+1}\right)=\prod _{k=1}^{n}sin\left(\frac{k\pi }{2n+1}\right)bych.n+1-j=k\Rightarrow$$$$2n+1={\left(2^n\right)}^2{\left(\prod _{k=1}^{n}sin\left(\frac{k\pi }{2n+1}\right)\right)}^2\Rightarrow$$$$\prod _{k=1}^{n}sin\left(\frac{k\pi }{2n+1}\right)=\frac{\sqrt{2n+1}}{2^n}.$$$$$$

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