IME / ITA ⇒ (ITA - 1975) Arco Metade Tópico resolvido

[MORANGA]

Seja senx= m-n/m+n , sendo m e n maior que zero. Determine o valor de tg( pi/4 - x/2) .

[jedi]

[tex3]\sen(x)=\frac{m-n}{m+n}[/tex3]

[tex3]cos^2(x)=1-\left(\frac{m-n}{m+n}\right)^2[/tex3]

[tex3]cos(x)=\frac{2\sqrt{mn}}{m+n}[/tex3]

[tex3]\cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}[/tex3]

[tex3]\cos(\frac{x}{2})=\sqrt{\frac{m+n+2\sqrt{mn}}{2(m+n)}}[/tex3]

[tex3]\sen^2(\frac{x}{2})=1-\sqrt{\frac{m+n+2\sqrt{mn}}{2(m+n)}}^2[/tex3]

[tex3]\sen(\frac{x}{2})=\sqrt{\frac{m+n-2\sqrt{mn}}{2(m+n)}}[/tex3]

[tex3]\tan\frac{x}{2}=\frac{\sqrt{\frac{m+n-2\sqrt{mn}}{2(m+n)}}}{\sqrt{\frac{m+n+2\sqrt{mn}}{2(m+n)}}}[/tex3]

[tex3]\tan\frac{x}{2}=\frac{\sqrt{m+n-2\sqrt{mn}}}{\sqrt{m+n+2\sqrt{mn}}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{\tan(\frac{\pi}{4})-\tan(\frac{x}{2})}{1+\tan(\frac{\pi}{4})\tan(\frac{x}{2})}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{1-\frac{\sqrt{m+n-2\sqrt{mn}}}{\sqrt{m+n+2\sqrt{mn}}}}{1+\frac{\sqrt{m+n-2\sqrt{mn}}}{\sqrt{m+n+2\sqrt{mn}}}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{\left(1-\frac{\sqrt{m+n-2\sqrt{mn}}}{\sqrt{m+n+2\sqrt{mn}}}\right)^2}{1-\frac{m+n-2\sqrt{mn}}{m+n+2\sqrt{mn}}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{1-2.\frac{\sqrt{m+n-2\sqrt{mn}}}{\sqrt{m+n+2\sqrt{mn}}}+\frac{m+n-2\sqrt{mn}}{m+n+2\sqrt{mn}}}{\frac{4\sqrt{mn}}{m+n+2\sqrt{mn}}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{\frac{2m+2n-2.\sqrt{(m+n)^2-(2\sqrt{mn})^2}}{m+n+2\sqrt{mn}}}{\frac{4\sqrt{mn}}{m+n+2\sqrt{mn}}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{2m+2n-2.\sqrt{m^2-2mn+n^2}}{4\sqrt{mn}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{2m+2n-2.\sqrt{(m-n)^2}}{4\sqrt{mn}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\frac{4n}{4\sqrt{mn}}[/tex3]

[tex3]\tan(\frac{\pi}{4}-\frac{x}{2})=\sqrt\frac{n}{m}[/tex3]

[MORANGA]

Pelo gabarito diz que é a resposta E, ou seja negativo.-√(n/m)

[Radius]

[tex3]\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}{\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}=\frac{\frac{1}{\sqrt{2}}\cdot \left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\frac{1}{\sqrt{2}}\cdot \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}[/tex3]

[tex3]\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}[/tex3]

multiplicando em cima e embaixo por [tex3]\cos \frac{x}{2}-\sin \frac{x}{2}[/tex3] :

[tex3]\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{\cos^2 \frac{x}{2}-2\cdot \sin \frac{x}{2}\cdot \cos \frac{x}{2}+\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}[/tex3]

[tex3]\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1-\sin x}{\cos x}[/tex3]

[tex3]\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1-\frac{m-n}{m+n}}{\pm \frac{2\sqrt{mn}}{m+n}}=\frac{2n}{\pm 2\sqrt{mn}}=\pm \sqrt{\frac{n}{m}}[/tex3]

[AndreBRasera]

Lindíssima resolução, Radius!