[SOLVED] Having trouble with this identity

Feb 2010
6
0
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)
 
Feb 2010
6
0
Also I'm only supposed to manipulate a single side of the equation.
 

skeeter

MHF Helper
Jun 2008
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North Texas
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 ...
 
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Reactions: Lucian
Feb 2010
6
0
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)