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📅 March 12, 2026 👤 scraface

Mar 12, 2026

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\sin 3x = 3\sin x - 4\sin^3 x
Let $ u = \sin x \in [-1, 1] $. Then:
\sin 3x \sin x = (3u - 4u^3)u = 3u^2 - 4u^4
Define $ g(u) = 3u^2 - 4u^4 $, $ u \in [-1, 1] $.
g'(u) = 6u - 16u^3 = 2u(3 - 8u^2)
Set $ g'(u) = 0 $: $ u = 0 $ or $ 3 - 8u^2 = 0 \Rightarrow u^2 = rac{...
Evaluate $ g(u) $:
At $ u = 0 $: $ g = 0 $
At $ u = \pm rac{\sqrt{6}}{4} $:
g = 3\left(rac{6}{16}

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