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how to get rid of water inside the ear - Decision Point

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 The argument of $ z_1 \overline{z_2} $ is $ heta_1 - heta_2 $ (modulo $ 360^\circ $), and the smallest positive argument is given to be $ 60^\circ $. Since cosine determines the angle uniquely within $ [0^\circ, 180^\circ] $ for this context (as it's the smallest angle between them), we have

📌 heta_1 - heta_2 \equiv 60^\circ \pmod{180^\circ},

📌 and since the angle is acute and positive, $ heta_1 - heta_2 = 60^\circ $ (modulo $ 360^\circ $, but the minimal positive value is taken). Therefore,

📌 \cos( heta_1 - heta_2) = \cos 60^\circ = rac{1}{2}.

📌 oxed{rac{1}{2}}

📌 Question: A museum curator is aligning two historical astrolabes represented by vectors $ egin{pmatrix} \cos lpha \ \sin lpha \end{pmatrix} $ and $ egin{pmatrix} \cos eta \ \sin eta \end{pmatrix} $ in the plane. If the cosine of the angle between them is $ rac{1}{3} $, and $ lpha + eta = 90^\circ $, find $ \sin(2lpha) $.

📌 The angle between two unit vectors $ \mathbf{u} $ and $ \mathbf{v} $ is given by $ \cos heta = \mathbf{u} \cdot \mathbf{v} $. So,

📌 \cos(lpha - eta) = rac{1}{3}.

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