Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

thank you sir!

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

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

thank you

$${thank}\:{you} \\ $$

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

thank you

$${thank}\:{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com