Question

(MAPOFEI-75) Demonstrar a identidade:

$\mathsf{tg(\frac{\pi}{4}+x)\cdot cotg(\frac{\pi}{4}-x)=\frac{1+sin(2x)}{1-sin(2x)}}$

Answer 1

Boa tarde Vinicius

formulas 

sen(a + b) = sen(a)*cos(b) + cos(a)*sen(b)
sen(a - b) = sen(a)*cos(b) - cos(a)*sen(b)

cos(a + b) = cos(a)*cos(b) - sen(a)*sen(b)
cos(a - b) = cos(a)*cos(b) + sen(a)*sen(b)

sen(π/4 + x) = sen(π/4)*cos(x) + cos(π/4)*sen(x) 
sen(π/4 - x) = sen(π/4)*cos(x) - cos(π/4)*sen(x) 

cos(π/4 + x) = cos(π/4)*cos(x) - sen(π/4)*sen(x)
cos(π/4 - x) = cos(π/4)*cos(x) + sen(π/4)*sen(x)

sen(π/4 + x) = √2/2*(cos(x) + sen(x))
sen(π/4 - x) = √2/2*(cos(x) - sen(x))

cos(π/4 + x) = √2/2*(cos(x) - sen(x))
cos(π/4 - x) = √2/2*(cos(x) + sen(x))

tg(π/4 + x) = (cos(x) + sen(x))/(cos(x) - sen(x))
tg(π/4 - x) = (cos(x) - sen(x))/(cos(x) + sen(x)) 

E = tg(π/4 + x) * cotg(π/4 - x) = tg(π/4 + x)/tg(π/4 - x) 

E = (cos(x) + sen(x))/(cos(x) - sen(x)) / (cos(x) - sen(x))/(cos(x) + sen(x)) 

E = (cos(x) + sen(x))²/(cos(x) - sen(x))² 

E = (cos²(x) + sen²(x) + 2sen(x)*cos(x))/(cos²(x) + sen²(x) - 2sen(x)*cos(x))

E = (1 + sen(2x)/(1 - sen(2x)) 

Answer 2

Olá Vinicius.


Precisamos conhecer as seguintes relações trigonométricas para resolver esse problema:


$\mathsf{tg(a+b)=\dfrac{sen(a+b)}{cos(a+b)}=\dfrac{sen(a)\cdot cos(b) + sen (b)\cdot cos(a)}{cos(a)\cdot cos(b)-sen(a)\cdot sen(b)}}\\\\\\\\\mathsf{cotg(a-b)=\dfrac{1}{tg(a-b)}=\dfrac{cos(a-b)}{sen(a-b)}=\dfrac{cos(a)\cdot cos(b)+sen(a)\cdot sen(b)}{sen(a)\cdot cos(b) - sen(b)\cdot cos(a)}}$

Resolvendo:

$\mathsf{\pi=180\cdot\Big(\dfrac{1}{4}\Big)\Rightarrow \dfrac{\pi}{4}=45}\\\\\mathsf{sen(45)=cos(45)=\dfrac{\sqrt{2}}{2}}\\\\\\\mathsf{tg\Big(\dfrac{\pi}{4}+x\Big)\cdot cotg\Big(\dfrac{\pi}{4}-x\Big)=\dfrac{1+sen(2x)}{1-sen(2x)}}\\\\=$
$\mathsf{\!\Big(\dfrac{sen(45)\cdot cos(x) + sen (x)\cdot cos(45)}{cos(45)\cdot cos(x)-sen(45)\cdot sen(x)}\Big)\!\cdot\!\Big(\dfrac{cos(45)\cdot cos(x)+sen(45)\cdot sen(x)}{sen(45)\cdot cos(x) - sen(x)\cdot cos(45)}\Big)}\\\\=$
$\mathsf{\Big(\dfrac{\frac{\sqrt{2}}{2}\cdot cos(x)+sen(x)\cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}\cdot cos(x)-\frac{\sqrt{2}}{2}\cdot sen(x)}\Big)\cdot\Big(\dfrac{\frac{\sqrt{2}}{2}\cdot cos(x)+\frac{\sqrt{2}}{2}\cdot sen(x)}{\frac{\sqrt{2}}{2}\cdot cos(x)-sen(x)\cdot \frac{\sqrt{2}}{2}}\Big)}\\\\=$
$\mathsf{\Big(\dfrac{\frac{\sqrt{2}}{2}\cdot[cos(x)+sen(x)]}{\frac{\sqrt{2}}{2}\cdot[cos(x)-sen(x)]}\Big)\cdot\Big(\dfrac{\frac{\sqrt{2}}{2}\cdot[cos(x)+sen(x)]}{\frac{\sqrt{2}}{2}\cdot[cos(x)-sen(x)]}\Big)\div\Big(\dfrac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\Big)}\\\\=$

$\mathsf{\dfrac{[cos(x)+sen(x)]^2}{[cos(x)-sen(x)]^2}=\dfrac{cos^2(x)+2\cdot cos(x)\cdot sen(x)+sen^2(x)}{cos^2(x)-2\cdot cos(x)\cdot sen(x) +sen^2(x)}}\\\\=\\\\\mathsf{cos^2(x)+sen^2(x)=1\Rightarrow cos^2(x)=1-sen^2(x)}\\\\=\\\\\mathsf{2\cdot cos(x)\cdot sen(x)=sen(2x)}\\\\=\\\\\mathsf{\dfrac{1-\diagup\!\!\!\!\!sen^2(x)+sen(2x)+\diagup\!\!\!\!\!sen^2(x)}{1-\diagup\!\!\!\!\!sen^2(x)-sen(2x)+\diagup\!\!\!\!sen^2(x)}}\\\\=\\\\\boxed{\mathsf{\dfrac{1+sen(2x)}{1-sen(2x)}}}~~~\checkmark$


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