Question

Sabendo que tg x=3/5, com x pertencente ao primeiro quadrante, determine tg (pi/4 + 2x)
Gabarito: -31/17

Answer 1

tang(x+y)=(tagx+tagy)/(1-tagx*tagx)

Tang pi/4=1

tag 2x=tag (x+x)=2tagx/1-tagx^2
=
=2×(3/5)/1-9/25=(6/5)/16/25=6/5×25/16=15/8

tag(pi/4+2x)=(1+15/8)/(1-(1×15/8)=(23/8)/(1-15/8)

=(23/8)/(-7/8)=-23/7.
achei uma resposta diferente do gabarito -23/7

Answer 2

Resposta:  − 23/7  (creio que haja algum erro no gabarito).

Explicação passo a passo:

Nesta tarefa, vamos utilizar a fórmula para a tangente da soma:

    $\mathsf{tg(\alpha+\beta)=\dfrac{tg\,\alpha+tg\,\beta}{1-tg\,\alpha\cdot tg\,\beta}}$

Primeiro, vamos calcular tg(2x):

    $\mathsf{tg(2x)=tg(x+x)}\\\\ \mathsf{tg(2x)=\dfrac{tg\,x+tg\,x}{1-tg\,x\cdot tg\,x}}\\\\\\ \mathsf{tg(2x)=\dfrac{2\,tg\,x}{1-tg^2\,x}}\\\\\\ \mathsf{tg(2x)=\dfrac{2\cdot \frac{3}{5}}{1-(\frac{3}{5})^2}}\\\\\\ \mathsf{tg(2x)=\dfrac{\frac{6}{5}}{1-\frac{9}{25}}}\\\\\\ \mathsf{tg(2x)=\dfrac{\frac{6}{5}}{~\frac{25-9}{25}~}}\\\\\\ \mathsf{tg(2x)=\dfrac{\frac{6}{5}}{~\frac{16}{25}~}}$

    $\mathsf{tg(2x)=\dfrac{6}{5}\cdot \dfrac{25}{16}}\\\\\\ \mathsf{tg(2x)=\dfrac{\diagup\!\!\!\! 2\cdot 3}{\diagdown\!\!\!\! 5}\cdot \dfrac{\diagdown\!\!\!\! 5\cdot 5}{\diagup\!\!\!\! 2\cdot 8}}\\\\\\ \mathsf{tg(2x)=\dfrac{3\cdot 5}{8}}\\\\\\ \mathsf{tg(2x)=\dfrac{15}{8}\qquad\checkmark}$

Finalmente, podemos calcular

    $\mathsf{tg\Big(\dfrac{\pi}{4}+2x\Big)=\dfrac{tg\,\frac{\pi}{4}+tg(2x)}{1-tg\,\frac{\pi}{4}\cdot tg(2x)}}\\\\\\\mathsf{tg\Big(\dfrac{\pi}{4}+2x\Big)=\dfrac{1+\frac{15}{8}}{1-1\cdot \frac{15}{8}}}\\\\\\ \mathsf{tg\Big(\dfrac{\pi}{4}+2x\Big)=\dfrac{\frac{8+15}{\diagup\!\!\!\! 8}}{~\frac{8-15}{\diagup\!\!\!\! 8}~}}\\\\\\ \mathsf{tg\Big(\dfrac{\pi}{4}+2x\Big)=\dfrac{8+15}{8-15}}\\\\\\ \mathsf{tg\Big(\dfrac{\pi}{4}+2x\Big)=\dfrac{23}{-7}}$

    $\mathsf{tg\Big(\dfrac{\pi}{4}+2x\Big)=-\,\dfrac{23}{7}\quad\longleftarrow\quad resposta.}$

Bons estudos! :-)