ITA 2024 — 2ª Fase — Questão 10 — Trigonometria
Sabendo que \(\tan(\alpha+\beta)=-2\) e \(\,\sin(\alpha)=(4\sqrt{5})\,\sin(\beta)\) para \(\alpha,\beta\in(0,\pi/2)\), calcule:
\[
\tan\!\left(\frac{\alpha+\beta}{2}\right),\qquad
\frac{\tan\!\left(\frac{\alpha-\beta}{2}\right)}{\tan\!\left(\frac{\alpha+\beta}{2}\right)},\qquad
\tan\!\left(\frac{\alpha-\beta}{2}\right).
\]
👀 Solução passo a passo
1) Cálculo de \(\tan\big(\tfrac{\alpha+\beta}{2}\big)\)
Seja \(\theta=\tfrac{\alpha+\beta}{2}\). Como \(\tan(\alpha+\beta)=-2\), \[ \tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}=-2 \;\;\Longrightarrow\;\; 2\tan\theta=-2(1-\tan^2\theta) \;\Longrightarrow\; \tan^2\theta-\tan\theta-1=0. \] Logo, \[ \tan\theta=\frac{1\pm\sqrt{5}}{2}. \] Como \(\alpha,\beta\in(0,\pi/2)\Rightarrow\alpha+\beta\in(\tfrac{\pi}{2},\pi)\Rightarrow \theta\in(\tfrac{\pi}{4},\tfrac{\pi}{2})\) e \(\tan\theta>1\), fica \[ \boxed{\displaystyle \tan\!\left(\frac{\alpha+\beta}{2}\right)=\frac{1+\sqrt{5}}{2}}. \]2) Relação entre meias-somas e meias-diferenças
Usando \[ \sin\alpha+\sin\beta=2\sin\!\left(\frac{\alpha+\beta}{2}\right)\cos\!\left(\frac{\alpha-\beta}{2}\right), \] \[ \sin\alpha-\sin\beta=2\sin\!\left(\frac{\alpha-\beta}{2}\right)\cos\!\left(\frac{\alpha+\beta}{2}\right), \] e dividindo a segunda pela primeira: \[ \frac{\sin\alpha-\sin\beta}{\sin\alpha+\sin\beta} = \frac{\tan\!\left(\frac{\alpha-\beta}{2}\right)}{\tan\!\left(\frac{\alpha+\beta}{2}\right)}. \] Como \(\sin\alpha=(4\sqrt{5})\sin\beta\), \[ \frac{(4\sqrt{5}-1)\sin\beta}{(4\sqrt{5}+1)\sin\beta} = \frac{4\sqrt{5}-1}{4\sqrt{5}+1} = \frac{3-\sqrt{5}}{5-\sqrt{5}} = \boxed{\displaystyle \frac{5-\sqrt{5}}{10}}. \] Portanto, \[ \boxed{\displaystyle \frac{\tan\!\left(\frac{\alpha-\beta}{2}\right)} {\tan\!\left(\frac{\alpha+\beta}{2}\right)} =\frac{5-\sqrt{5}}{10}}. \]3) Cálculo de \(\tan\big(\tfrac{\alpha-\beta}{2}\big)\)
Multiplicando o resultado anterior por \(\tan\big(\tfrac{\alpha+\beta}{2}\big)=\frac{1+\sqrt{5}}{2}\): \[ \tan\!\left(\frac{\alpha-\beta}{2}\right) =\frac{5-\sqrt{5}}{10}\cdot\frac{1+\sqrt{5}}{2} =\frac{(5-\sqrt{5})(1+\sqrt{5})}{20} =\frac{4\sqrt{5}}{20} =\boxed{\displaystyle \frac{\sqrt{5}}{5}}. \]
Seja \(\theta=\tfrac{\alpha+\beta}{2}\). Como \(\tan(\alpha+\beta)=-2\), \[ \tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}=-2 \;\;\Longrightarrow\;\; 2\tan\theta=-2(1-\tan^2\theta) \;\Longrightarrow\; \tan^2\theta-\tan\theta-1=0. \] Logo, \[ \tan\theta=\frac{1\pm\sqrt{5}}{2}. \] Como \(\alpha,\beta\in(0,\pi/2)\Rightarrow\alpha+\beta\in(\tfrac{\pi}{2},\pi)\Rightarrow \theta\in(\tfrac{\pi}{4},\tfrac{\pi}{2})\) e \(\tan\theta>1\), fica \[ \boxed{\displaystyle \tan\!\left(\frac{\alpha+\beta}{2}\right)=\frac{1+\sqrt{5}}{2}}. \]2) Relação entre meias-somas e meias-diferenças
Usando \[ \sin\alpha+\sin\beta=2\sin\!\left(\frac{\alpha+\beta}{2}\right)\cos\!\left(\frac{\alpha-\beta}{2}\right), \] \[ \sin\alpha-\sin\beta=2\sin\!\left(\frac{\alpha-\beta}{2}\right)\cos\!\left(\frac{\alpha+\beta}{2}\right), \] e dividindo a segunda pela primeira: \[ \frac{\sin\alpha-\sin\beta}{\sin\alpha+\sin\beta} = \frac{\tan\!\left(\frac{\alpha-\beta}{2}\right)}{\tan\!\left(\frac{\alpha+\beta}{2}\right)}. \] Como \(\sin\alpha=(4\sqrt{5})\sin\beta\), \[ \frac{(4\sqrt{5}-1)\sin\beta}{(4\sqrt{5}+1)\sin\beta} = \frac{4\sqrt{5}-1}{4\sqrt{5}+1} = \frac{3-\sqrt{5}}{5-\sqrt{5}} = \boxed{\displaystyle \frac{5-\sqrt{5}}{10}}. \] Portanto, \[ \boxed{\displaystyle \frac{\tan\!\left(\frac{\alpha-\beta}{2}\right)} {\tan\!\left(\frac{\alpha+\beta}{2}\right)} =\frac{5-\sqrt{5}}{10}}. \]3) Cálculo de \(\tan\big(\tfrac{\alpha-\beta}{2}\big)\)
Multiplicando o resultado anterior por \(\tan\big(\tfrac{\alpha+\beta}{2}\big)=\frac{1+\sqrt{5}}{2}\): \[ \tan\!\left(\frac{\alpha-\beta}{2}\right) =\frac{5-\sqrt{5}}{10}\cdot\frac{1+\sqrt{5}}{2} =\frac{(5-\sqrt{5})(1+\sqrt{5})}{20} =\frac{4\sqrt{5}}{20} =\boxed{\displaystyle \frac{\sqrt{5}}{5}}. \]
Respostas:
\(\displaystyle \tan\!\left(\frac{\alpha+\beta}{2}\right)=\frac{1+\sqrt{5}}{2},\quad
\frac{\tan\!\left(\frac{\alpha-\beta}{2}\right)}{\tan\!\left(\frac{\alpha+\beta}{2}\right)}=\frac{5-\sqrt{5}}{10},\quad
\tan\!\left(\frac{\alpha-\beta}{2}\right)=\frac{\sqrt{5}}{5}.\)
