I-would-like-that-you-help-me-to-show-this-equality-16cos-24-cos-5-24-cos-7-24-cos-11-24-1-

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

I would like that you help me to show this equality: 16cos (Π/(24))cos((5Π)/(24))cos((7Π)/(24))cos((11Π)/(24))=1

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

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

identite used sin(a)=cos((π/2)−a) sin((π/2)+α)=cos(α), sin(2x)=2sin(x)cos(x)

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

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

cos(((11π)/(24)))=sin((π/(24))) cos(((5π)/(24)))=sin(((7π)/(24))) 16cos((π/(24)))cos(((5π)/(24)))cos(((7π)/(24)))cos(((11π)/(24)))=16cos((π/(24)))sin((π/(24)))cos(((7π)/(24)))sin(((7π)/(24))) =4sin((π/(12)))sin(((7π)/(12))) =4sin((π/(12)))sin((π/2)+(π/(12)))=2.2sin((π/(12)))cos((π/(12))) =2sin((π/6))=2.(1/2)=1

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

Commented by mathocean1 last updated on 06/Dec/19