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Question Number 207065 by manxsol last updated on 05/May/24

$$$$$$f\left(x\right)=[cos2x+cos3x][cos4x+cos6x][[cosx+cos5x]$$$$evaluarf\left(\frac{2\pi }{13}\right)$$

Answered by Berbere last updated on 06/May/24

$$4cos\left(x\right)cos\left(5x\right)cos\left(2x\right)cos\left(3x\right).[cos\left(2x\right)+cos\left(3x\right)]]a2($$$$8cos\left(\frac{\pi }{13}\right)cos\left(\frac{5\pi }{13}\right)cos\left(\frac{2\pi }{13}\right)cos\left(\frac{4\pi }{13}\right)cos\left(\frac{6\pi }{13}\right)cos\left(\frac{10\pi }{13}\right)$$$$=f\left(\frac{\pi }{12}\right)-8\underset{k=1}{\prod ^6}cos\left(\frac{k\pi }{13}\right)=-8\underset{k=7}{\sum ^{12}}cos\left(k\frac{\pi }{13}\right)$$$$\Rightarrow {\left(\underset{k=1}{\prod ^{12}}\left(cos\left(k\frac{\pi }{13}\right)\right)\right)}^{\frac{1}{2}}=\left(\underset{k=1}{\prod ^6}cos\left(\frac{k\pi }{13}\right)\right)$$$$=p\left(z\right)=Z^{13}-1=\underset{k=0}{\prod ^{12}}(z-e^{\frac{2ik\pi }{13}})=p(-1)=-2$$$$=-\underset{k=0}{\prod ^{12}}(1+e^{\frac{2ik\pi }{13}})=-2^{13}\underset{k=0}{\prod ^{12}}cos\left(\frac{k\pi }{13}\right).e^{i12\pi }=-2$$$$=\underset{k=0}{\prod ^{12}}cos\left(k\frac{\pi }{13}\right)=\frac{1}{2^{12}}\Rightarrow f\left(\frac{\pi }{12}\right)=-2^3.\frac{1}{2^6}=-\frac{1}{8}$$$$$$

Commented by manxsol last updated on 07/May/24

$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{sir}\mathrm{it}\mathrm{is}\mathrm{complicatedt}$$$$\mathrm{wha}\mathrm{theory}\mathrm{do}\mathrm{you}\mathrm{need}\mathrm{to}\mathrm{study}\mathrm{tou}$$$$\mathrm{nderstand}\mathrm{it}$$

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