I've been at this for a little while now but I'm not quite sure how to proceed. Would someone please help me? Cos3(theta) = 4Cos^3(theta) - 3Cos(theta)
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Also I'm only supposed to manipulate a single side of the equation.
Originally Posted by Lucian I've been at this for a little while now but I'm not quite sure how to proceed. Would someone please help me? Cos3(theta) = 4Cos^3(theta) - 3Cos(theta) $\displaystyle \cos(3t) = $ $\displaystyle \cos(2t+t) =$ $\displaystyle \cos(2t)\cos(t) - \sin(2t)\sin(t) =$ $\displaystyle [2\cos^2(t) - 1]\cos(t) - 2\sin^2(t)\cos(t) =$ $\displaystyle [2\cos^2(t) - 1]\cos(t) - 2[1 - \cos^2(t)]\cos(t) =$ finish it ...
I think I've got it, thank you for the help Skeeter! (2Cos^3(t)-Cos(t)) - (2)(Cos(t)-Cos^3(t)) (2Cos^3(t)-Cos(t)) - (2Cos(t) - 2Cos^3(t)) 2Cos^3(t) -Cos(t) - 2Cos(t) + 2Cos^3(t) 4Cos^3(t) - 3Cos(t)
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