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ITA 2021 — 2ª Fase — Questão 9 — Trigonometria
Sejam \(\alpha,\beta,\gamma\in\mathbb{R}\) tais que \(\alpha+\beta+\gamma=3\pi\),
\(\sin\alpha+\sin\beta+\sin\gamma=\dfrac12\) e \(\cos\alpha+\cos\beta+\cos\gamma=-\dfrac12\).
Determine o valor de \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma \).
\(\sin\alpha+\sin\beta+\sin\gamma=\dfrac12\) e \(\cos\alpha+\cos\beta+\cos\gamma=-\dfrac12\).
Determine o valor de \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma \).
👀 Solução passo a passo
Denote \(S_{\sin}=\sin\alpha+\sin\beta+\sin\gamma=\dfrac12\) e
\(S_{\cos}=\cos\alpha+\cos\beta+\cos\gamma=-\dfrac12\).
Considere a diferença dos quadrados:
\[
S_{\cos}^2-S_{\sin}^2
=\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)
-\big(\sin^2\alpha+\sin^2\beta+\sin^2\gamma\big)
+2\!\sum_{\text{cíc}}\!(\cos\alpha\cos\beta-\sin\alpha\sin\beta).
\]
Como \(\cos^2\theta-\sin^2\theta=\cos 2\theta\) e
\(\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta)\),
obtemos
\[
S_{\cos}^2-S_{\sin}^2
=\sum_{\text{cíc}}\cos 2\theta
+2\!\sum_{\text{cíc}}\cos(\alpha+\beta).
\]
Mas \(S_{\cos}^2-S_{\sin}^2=\left(-\dfrac12\right)^2-\left(\dfrac12\right)^2=0\).
E, como \(\alpha+\beta+\gamma=3\pi\),
\(\cos(\alpha+\beta)=\cos(3\pi-\gamma)=-\cos\gamma\), logo
\[
\sum_{\text{cíc}}\cos(\alpha+\beta)=-(\cos\alpha+\cos\beta+\cos\gamma)=-\left(-\frac12\right)=\frac12.
\]
Portanto
\[
\sum_{\text{cíc}}\cos 2\theta + 2\cdot\frac12=0
\ \Longrightarrow\
\sum_{\text{cíc}}\cos 2\theta=-1.
\]
Usando \(\cos 2\theta=2\cos^2\theta-1\),
\[
\sum_{\text{cíc}}(2\cos^2\theta-1)=-1
\ \Longrightarrow\
2\!\sum_{\text{cíc}}\cos^2\theta-3=-1
\ \Longrightarrow\
\sum_{\text{cíc}}\cos^2\theta=1.
\]
Resposta: \(\boxed{1}\).
🔗 Veja também a questão anterior:
Matemática ITA 2021 — 2ª Fase — Questão 8
