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Serie B players - AMAZONAWS
Serie B players - AMAZONAWS
📅 March 11, 2026
👤 scraface
Mar 11, 2026
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📌 \[ \Rightarrow \quad \cos\left(\frac{\pi}{2} + \phi\right) = 0 \]
📌 Solution: Expand $ (\sin x + 2\cos x)^2 = \sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Add $ \sin^2 x $: total $ 2\sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Simplify using identities: $ 2(1 - \cos^2 x) + 2\sin 2x + 4\cos^2 x = 2 + 2\cos^2 x + 2\sin 2x $. Let $ u = \cos^2 x $, $ \sin 2x = 2\sin x \cos x $. Alternatively, rewrite original expression as $ \sin^2 x + 4\sin x \cos x + 4\cos^2 x + \sin^2 x = 2\sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Let $ f(x) = 2\sin^2 x + 4\sin x \cos x + 4\cos^2 x $. Use $ \sin^2 x = rac{1 - \cos 2x}{2} $, $ \cos^2 x = rac{1 + \cos 2x}{2} $, $ \sin x \cos x = rac{\sin 2x}{2} $:
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