Question

Determine n tal que (1 + tg(1o ))(1 + tg(2o )...(1 + tg(45o )) = 2n .

Answer 1

$\displaystyle \sf (1+tg\ 1^\circ)\cdot (1+tg\ 2^\circ)\cdot(1+tg\ 3^\circ)...(1+tg\ 45^\circ)=2^n$

Vamos estudar o que acontece quando fazemos uma multiplicação desse tipo :

$\displaystyle \sf (1+tg\ \alpha )\cdot (1+tg\ \beta ) = 1+tg\ \alpha+ tg\ \beta +tg\ \alpha \cdot tg\ \beta \\\\\\ \underline{Vamos \ usar\ o \ arco\ soma\ da\ tangente} : \\\\ tg(\alpha+\beta)=\frac{tg\ \alpha+tg\ \beta}{1-tg\ \alpha\cdot tg\ \beta} \\\\\\ tg\ \alpha + tg\ \beta = tg(\alpha+\beta)\cdot [1-tg\ \alpha\cdot tg\ \beta] \\\\\\ \underline{vamos\ subsituir \ essa\ soma\ de \ tangentes\ de \ onde\ paramos} :$

$\sf (1+tg\ \alpha )\cdot (1+tg\ \beta ) = 1\+\underbrace{\sf tg\ \alpha+ tg\ \beta}_{ \sf tg(\alpha+\beta)\cdot[1-tg\ \alpha\cdot tg\ \beta] } +tg\ \alpha \cdot tg\ \beta$

$\sf (1+tg\ \alpha )\cdot (1+tg\ \beta ) = 1+tg(\alpha+\beta)\cdot[1-tg\ \alpha\cdot tg\ \beta]+tg\ \alpha \cdot tg\ \beta \\\\\\ (1+tg\ \alpha )\cdot (1+tg\ \beta )=1+tg(\alpha+\beta)-tg(\alpha+\beta)\cdot tg\ \alpha\cdot tg\ \beta+tg\ \alpha\cdot tg\ \beta\\\\\\ (1+tg\ \alpha )\cdot (1+tg\ \beta )= 1+tg(\alpha+\beta)+tg\ \alpha\cdot tg\ \beta\cdot [1-tg(\alpha+\beta)]$

Então se pegarmos angulos, cujo

$\sf tg(\alpha+\beta)=1$

$\alpha+ \beta=45^\circ$

teremos que :

$\sf (1+tg\ \alpha )\cdot (1+tg\ \beta )= 1+\underbrace{tg(\alpha+\beta)}_{\displaystyle 1 }+tg\ \alpha\cdot tg\ \beta\cdot [1-\underbrace{tg(\alpha+\beta)}_{\displaystyle 1 }]\\\\ (1+tg\ \alpha )\cdot (1+tg\ \beta )= 1+1+tg\ \alpha\cdot tg\ \beta\cdot [1-1] \\\\ (1+tg\ \alpha )\cdot (1+tg\ \beta )= 2+tg\ \alpha\cdot tg\ \beta\cdot [0] \\\\\\ \boxed{\sf (1+tg\ \alpha )\cdot (1+tg\ \beta )=2 }$

Voltando para a questão, vamos pegar os produtos cujo os angulos somam 45º, assim :

$\sf (1+tg\ 1^\circ)\cdot(1+tg\ 44^{\circ})\cdot(1+tg\ 2^\circ)\cdot(1+tg \ 43^\circ)....(1+tg\ 45^\circ) = 2^n \\\\\\ \underbrace{\sf (1+tg\ 1^\circ)\cdot(1+tg\ 44^{\circ})}_{2}\cdot \underbrace{\sf (1+tg\ 2^\circ)\cdot(1+tg\ 43^{\circ})}_{2}....(1+1)=2^n \\\\\\ \underbrace{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2....}_{\displaystyle 22 \ vezes}2=2^n \\\\\\ 2^{22}\cdot 2 = 2^n \\\\ 2^{23}=2^n \\\\ \underline{Portanto}} : \\\\ \huge\boxed{\displaystyle \sf n = 23 }\checkmark$