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Hard Trigonometric identity

For all \({\alpha},{\beta},{\gamma}\) show that \(\sum_{cyclic}\sin^3{\alpha}\sin(\beta-\gamma)=-\sin(\beta-\gamma)\sin(\gamma-\alpha)\sin(\alpha-\beta)\times\sin(\alpha+\beta+\gamma)\)

Note by Harry Jones
1 month, 3 weeks ago

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@Harry Jones Use SIN 3A formula Using this formula u get substitution for sin ^3 alpha Neel Khare · 1 week, 3 days ago

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Any solutions? Harry Jones · 2 weeks, 3 days ago

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