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Question Number 197327 by mnjuly1970 last updated on 13/Sep/23

$$$$$$trigonometry...$$$$$$$$\mathrm{P}=\underset{k=1}{\prod ^{44}}(1+tan\left(k\right))=?$$$$$$$$$$

Answered by witcher3 last updated on 13/Sep/23

$$\mathrm{\Pi }\frac{\sin \left(\mathrm{k}\right)+\cos \left(\mathrm{k}\right)}{\cos \left(\mathrm{k}\right)}=\frac{\mathrm{\Pi }\left(\sqrt{2}\sin (\mathrm{k}+45)\right)}{\mathrm{\Pi }\cos \left(\mathrm{k}\right)}$$$$\underset{\mathrm{k}=1}{\prod ^{44}}\sqrt{2}\sin (\mathrm{k}+45)=2^{22}\underset{1}{\coprod ^{44}}\sin (90-\mathrm{k})=2^{22}\mathrm{\Pi }\cos \left(\mathrm{k}\right)$$$$\mathrm{P}=2^{22}$$

Commented by mnjuly1970 last updated on 13/Sep/23

$$thanksalotsirWitcher$$

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

$$Iheard‘‘thequoqroductof{(90-k)}^{″}.$$$$Pleasereadtheblueone.$$

Answered by deleteduser1 last updated on 13/Sep/23

$$P=\underset{k=1}{\prod ^{44}}\frac{\sqrt{2}sin(k+45)}{cos\left(k\right)}=\frac{{\left(\sqrt{2}\right)}^{44}sin\left(46\right)sin\left(47\right)...sin\left(89\right)}{cos\left(1\right)cos\left(2\right)...cos\left(44\right)}$$$$Butsin\left(x\right)=cos(90-x)\Rightarrow \frac{sin\left(x\right)}{cos(90-x)}=1$$$$P=2^{22}\frac{sin\left(89\right)}{cos\left(1\right)}\times \frac{sin\left(88\right)}{cos\left(2\right)}\times ...\times \frac{sin\left(47\right)}{cos\left(43\right)}\times \frac{sin\left(46\right)}{cos\left(44\right)}$$$$=2^{22}\times 1$$

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