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Question Number 80929 by mathocean1 last updated on 08/Feb/20

$$\mathrm{show}\mathrm{that}$$$$cos\frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=-\frac{1}{8}$$

Commented by jagoll last updated on 08/Feb/20

$$letx=\frac{\pi }{7}$$$$\cos x\cos 2x\cos 4x\times \frac{2\sin x}{2\sin x}=$$$$\frac{\sin 2x\cos 2x\cos 4x}{2\sin x}=\frac{\sin 4x\cos 4x}{4\sin x}=$$$$\frac{\sin 8x}{8\sin x}=\frac{\sin \left(\frac{8\pi }{7}\right)}{8\sin \left(\frac{\pi }{7}\right)}=$$$$\frac{-\sin \left(\frac{\pi }{7}\right)}{8\sin \left(\frac{\pi }{7}\right)}=-\frac{1}{8}$$

Commented by mr W last updated on 08/Feb/20

$$saycos\frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=x$$$$2\sin \frac{\pi }{7}cos\frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=2\sin \frac{\pi }{7}\times x$$$$\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=2\sin \frac{\pi }{7}\times x$$$$2\sin \frac{2\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}=2\times 2\sin \frac{\pi }{7}\times x$$$$\sin \frac{4\pi }{7}\cos \frac{4\pi }{7}=4\sin \frac{\pi }{7}\times x$$$$2\sin \frac{4\pi }{7}\cos \frac{4\pi }{7}=2\times 4\sin \frac{\pi }{7}\times x$$$$\sin \frac{8\pi }{7}=8\sin \frac{\pi }{7}\times x$$$$\sin (\pi +\frac{\pi }{7})=8\sin \frac{\pi }{7}\times x$$$$-\sin \frac{\pi }{7}=8\sin \frac{\pi }{7}\times x$$$$-1=8x$$$$-\frac{1}{8}=x$$

Commented by jagoll last updated on 08/Feb/20

$$wawgreatsir$$

Commented by peter frank last updated on 08/Feb/20

$$thankyou$$

Commented by mathmax by abdo last updated on 08/Feb/20

$$letA=cos\left(\frac{\pi }{7}\right).cos\left(\frac{2\pi }{7}\right)cos\left(\frac{4\pi }{7}\right)$$$$A\times sin\left(\frac{\pi }{7}\right)=\frac{1}{2}sin\left(\frac{2\pi }{7}\right)cos\left(\frac{2\pi }{7}\right)cos\left(\frac{4\pi }{7}\right)=\frac{1}{4}sin\left(\frac{4\pi }{7}\right)cos\left(\frac{4\pi }{7}\right)$$$$=\frac{1}{8}sin\left(\frac{8\pi }{7}\right)=\frac{1}{8}sin(\pi +\frac{\pi }{7})=-\frac{1}{8}sin\left(\frac{\pi }{7}\right)bysimplification$$$$wegetA=-\frac{1}{8}$$

Answered by mind is power last updated on 09/Feb/20

$$z^7-1=0$$$$z^7=-1=e^{i\left(2k\right)\pi }$$$$z_k=e^{i\frac{\left(2k\right)\pi }{7}}$$$$z^7-1=(z-1)\underset{k=1}{\prod ^6}(z-z_k)$$$$=(z-1)\underset{k=1}{\prod ^{k=3}}(z-z_k)(z-z_k^{-})$$$$=\underset{k=1}{\prod ^{k=3}}(1-e^{i\frac{\left(2k\right)\pi }{7}})(1-e^{-\frac{i\left(2k\right)\pi }{7}})$$$$z=-1\Rightarrow$$$$1=2^6\underset{k=1}{\prod ^3}{cos}^2\left(\frac{k\pi }{7}\right)$$$$\Rightarrow \frac{1}{2^3}=cos\left(\frac{\pi }{7}\right)cos\left(\frac{2\pi }{7}\right)cos\left(\frac{3\pi }{7}\right)$$$$cos\left(\frac{3\pi }{7}\right)=-cos(\pi -\frac{3\pi }{7})=-cos\left(\frac{4\pi }{7}\right)\Rightarrow$$$$-cos\left(\frac{\pi }{7}\right)cos\left(\frac{2\pi }{7}\right)cos\left(\frac{4\pi }{7}\right)=\frac{1}{2^3}=\frac{1}{8}\Rightarrow$$$$cos\left(\frac{\pi }{7}\right)cos\left(\frac{2\pi }{7}\right)cos\left(\frac{4\pi }{7}\right)=-\frac{1}{8}$$

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