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Question Number 75048 by mathocean1 last updated on 06/Dec/19

$$\mathrm{I}\mathrm{would}\mathrm{like}\mathrm{that}\mathrm{you}\mathrm{help}\mathrm{me}\mathrm{to}$$$$\mathrm{show}\mathrm{this}\mathrm{equality}:$$$$16\cos \frac{\mathrm{\Pi }}{24}\cos \frac{5\mathrm{\Pi }}{24}\cos \frac{7\mathrm{\Pi }}{24}\cos \frac{11\mathrm{\Pi }}{24}=1$$

Commented by mind is power last updated on 06/Dec/19

$$\mathrm{identite}\mathrm{used}$$$$\sin \left(\mathrm{a}\right)=\cos (\frac{\pi }{2}-\mathrm{a})$$$$\sin (\frac{\pi }{2}+\alpha )=\cos \left(\alpha \right),\sin \left(2\mathrm{x}\right)=2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)$$

Answered by mind is power last updated on 06/Dec/19

$$\cos \left(\frac{11\pi }{24}\right)=\sin \left(\frac{\pi }{24}\right)$$$$\cos \left(\frac{5\pi }{24}\right)=\sin \left(\frac{7\pi }{24}\right)$$$$16\cos \left(\frac{\pi }{24}\right)\cos \left(\frac{5\pi }{24}\right)\cos \left(\frac{7\pi }{24}\right)\cos \left(\frac{11\pi }{24}\right)=16\cos \left(\frac{\pi }{24}\right)\sin \left(\frac{\pi }{24}\right)\cos \left(\frac{7\pi }{24}\right)\sin \left(\frac{7\pi }{24}\right)$$$$=4\sin \left(\frac{\pi }{12}\right)\sin \left(\frac{7\pi }{12}\right)$$$$=4\sin \left(\frac{\pi }{12}\right)\sin (\frac{\pi }{2}+\frac{\pi }{12})=2.2\sin \left(\frac{\pi }{12}\right)\cos \left(\frac{\pi }{12}\right)$$$$=2\sin \left(\frac{\pi }{6}\right)=2.\frac{1}{2}=1$$

Commented by mathocean1 last updated on 06/Dec/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!$$

Commented by peter frank last updated on 06/Dec/19

$$thankyou$$

Commented by peter frank last updated on 06/Dec/19

$$thankyou$$

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