Question Number 197327 by mnjuly1970 last updated on 13/Sep/23
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{trigonometry}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{P}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{44}} {\prod}}\left(\:\:\mathrm{1}\:+\:{tan}\left({k}\right)\:\right)\:=\:?\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Answered by witcher3 last updated on 13/Sep/23
$$\Pi\frac{\mathrm{sin}\left(\mathrm{k}\right)+\mathrm{cos}\left(\mathrm{k}\right)}{\mathrm{cos}\left(\mathrm{k}\right)}=\frac{\Pi\left(\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{k}+\mathrm{45}\right)\right)}{\Pi\mathrm{cos}\left(\mathrm{k}\right)} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{44}} {\prod}}\sqrt{\mathrm{2}}\mathrm{sin}\left(\mathrm{k}+\mathrm{45}\right)=\mathrm{2}^{\mathrm{22}} \underset{\mathrm{1}} {\overset{\mathrm{44}} {\coprod}}\mathrm{sin}\left(\mathrm{90}−\mathrm{k}\right)=\mathrm{2}^{\mathrm{22}} \Pi\mathrm{cos}\left(\mathrm{k}\right) \\ $$$$\mathrm{P}=\mathrm{2}^{\mathrm{22}} \\ $$
Commented by mnjuly1970 last updated on 13/Sep/23
$$\:\:\:\:{thanks}\:{alot}\:\:{sir}\:\:{Witcher} \\ $$
Commented by witcher3 last updated on 14/Sep/23
$$\mathrm{withe}\:\mathrm{Pleasur}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{You} \\ $$
Commented by MathematicalUser2357 last updated on 09/Jan/24
$${I}\:{heard}\:“{the}\:{quoqroduct}\:{of}\:\left(\mathrm{90}−{k}\right)''. \\ $$$${Please}\:{read}\:{the}\:{blue}\:{one}. \\ $$
Answered by AST last updated on 13/Sep/23
$${P}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{44}} {\prod}}\frac{\sqrt{\mathrm{2}}{sin}\left({k}+\mathrm{45}\right)}{{cos}\left({k}\right)}=\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{44}} {sin}\left(\mathrm{46}\right){sin}\left(\mathrm{47}\right)…{sin}\left(\mathrm{89}\right)}{{cos}\left(\mathrm{1}\right){cos}\left(\mathrm{2}\right)…{cos}\left(\mathrm{44}\right)} \\ $$$${But}\:{sin}\left({x}\right)={cos}\left(\mathrm{90}−{x}\right)\Rightarrow\frac{{sin}\left({x}\right)}{{cos}\left(\mathrm{90}−{x}\right)}=\mathrm{1} \\ $$$${P}=\mathrm{2}^{\mathrm{22}} \frac{{sin}\left(\mathrm{89}\right)}{{cos}\left(\mathrm{1}\right)}×\frac{{sin}\left(\mathrm{88}\right)}{{cos}\left(\mathrm{2}\right)}×…×\frac{{sin}\left(\mathrm{47}\right)}{{cos}\left(\mathrm{43}\right)}×\frac{{sin}\left(\mathrm{46}\right)}{{cos}\left(\mathrm{44}\right)} \\ $$$$=\mathrm{2}^{\mathrm{22}} ×\mathrm{1} \\ $$