Question

$\mathsf{Utilizando~a~rela\c{c}\~ao~sin~\dfrac{\pi}{n}=\dfrac{\ell_n}{2}~~(\ell_n=lado~do~pol\'igono~regular~de~n~lados}}\\\mathsf{inscrito~na~circunfer\^encia,~com~n\in\mathbb{N},~n \geq 3).}\\\\\\\mathsf{Calcule:~sin~\dfrac{\pi}{8},~cos~\dfrac{\pi}{8}~e~tg~\dfrac{\pi}{8}.}\\\\\\\mathsf{Demonstre~visualmente~o~c\'alculo~de~\ell_{n},~lado~do~pol\'igono~regular~inscrito.}$

Answer 1

O laTex não aceitou a continuação da digitação, segue resolução anexada.$\frac{ a_{4} }{r} =sen \frac{ \pi }{4} =\ \textgreater \ \frac{ a_{4} }{r}= \frac{ \sqrt{2} }{2} =\ \textgreater \ a_{4} = \frac{r \sqrt{2} }{2}= \frac{1 \sqrt{2} }{2} = \frac{ \sqrt{2} }{2} \\ \\ TP=1-r=\ \textgreater \ TP=1-OT=\ \textgreater \ TP=1- a_{4} \\ TP=1- \frac{ \sqrt{2} }{2} \\ l_{4}=TQ= \frac{r \sqrt{2} }{2} = \frac{1 \sqrt{2} }{2} = \frac{ \sqrt{2} }{2} \\ \\ NotriânguloPQT \\ PQ^{2} = TP^{2} + TQ^{2} =\ \textgreater \ TQ^{2} = (1- \frac{ \sqrt{2} }{2} )^{2} + ( \frac{2}{2}) ^{2} =\ \textgreater \ \\ TQ^{2} =1- \sqrt{2}+\frac{2}{4} + \frac{2}{4}=1- \sqrt{2} +1 = \\ 2-$