L Lucian Feb 2010 6 0 May 9, 2010 #1 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)

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)

L Lucian Feb 2010 6 0 May 9, 2010 #2 Also I'm only supposed to manipulate a single side of the equation.

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas May 9, 2010 #3 Lucian said: 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) Click to expand... \(\displaystyle \cos(3t) = \) \(\displaystyle \cos(2t+t) =\) \(\displaystyle \cos(2t)\cos(t) - \sin(2t)\sin(t) =\) \(\displaystyle [2\cos^2(t) - 1]\cos(t) - 2\sin^2(t)\cos(t) =\) \(\displaystyle [2\cos^2(t) - 1]\cos(t) - 2[1 - \cos^2(t)]\cos(t) =\) finish it ... Reactions: Lucian

Lucian said: 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) Click to expand... \(\displaystyle \cos(3t) = \) \(\displaystyle \cos(2t+t) =\) \(\displaystyle \cos(2t)\cos(t) - \sin(2t)\sin(t) =\) \(\displaystyle [2\cos^2(t) - 1]\cos(t) - 2\sin^2(t)\cos(t) =\) \(\displaystyle [2\cos^2(t) - 1]\cos(t) - 2[1 - \cos^2(t)]\cos(t) =\) finish it ...

L Lucian Feb 2010 6 0 May 9, 2010 #4 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)

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)