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Home » \sin(2lpha) = \cos(2lpha - 90^\circ) = rac13. - Appfinity Technologies

\sin(2lpha) = \cos(2lpha - 90^\circ) = rac13. - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Substitute \(
ho = c \sin(\phi)\) into the expressions:
x = c \sin(\phi) \cdot \sin(\phi) \cos heta = c \sin^2(\phi) \cos heta
y = c \sin^2(\phi) \sin heta
z = c \sin(\phi) \cos(\phi)
Now compute \(x^2 + y^2\):
x^2 + y^2 = c^2 \sin^4(\phi) (\cos^2 heta + \sin^2 heta) = c^2 \sin^4(...
From the identity \(\sin^2(\phi) = rac{
ho^2}{c^2}\), but express \(x^2 + y^2\) in terms of \(z\):
z = c \sin\phi \cos\phi = rac{c}{2} \sin(2\phi)

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