sin³(3x) = ½(sin(3x))(1-cos(6x)) I'm just stumped...everything I've tried has turned into a complete mess. Thanks a bunch for any ideas!
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Originally Posted by generalmeow sin³(3x) = ½(sin(3x))(1-cos(6x)) I'm just stumped...everything I've tried has turned into a complete mess. Thanks a bunch for any ideas! $\displaystyle \displaystyle \sin^3(3x)=\sin(3x)\sin^2(3x)=\sin(3x)\left[\frac{1}{2}(1-\cos(6x))\right]$
Originally Posted by generalmeow sin³(3x) = ½(sin(3x))(1-cos(6x)) I'm just stumped...everything I've tried has turned into a complete mess. Thanks a bunch for any ideas! For this one, show that RHS implies LHS. For a hint, reference the spoiler. Spoiler: $\displaystyle \sin^2(\alpha)=\tfrac{1}{2}(1-\cos(2\alpha))$ EDIT: TES beat me!
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