abstract algebra question

[tam14]

Hi,

I am trying to prove cos3(-) = 4 cos^3(-) - 3 cos(-) from Euler's formula e^(i(-)) = cos(-) + sin(-) where (-) is my theta. I don't know how to type the theta so that looks a little confusing. I think my problem is I am not that familiar with Euler's formula. Any help would be greatly appreciated!

Thanks![/img]

 

[pka]

\(\displaystyle \L
\begin{array}{l}
e^{i\theta } = \cos (\theta ) + i\sin (\theta )\quad \& \quad e^{ - i\theta } = \cos (\theta ) - i\sin (\theta ) \\
2\cos (\theta ) = e^{i\theta } + e^{ - i\theta } \\
\left( {2\cos (\theta )} \right)^3 = \left( {e^{i\theta } + e^{ - i\theta } } \right)^3 \\
8\cos ^3 (\theta ) = e^{i3\theta } + 3e^{i\theta } + 3e^{ - i\theta } + e^{ - i3\theta } \\
\\
\mbox{so} \\
\\
e^{i3\theta } + e^{ - i3\theta } = 2\cos (3\theta )\quad \& \quad 3e^{i\theta } + 3e^{ - i\theta } = 6\cos (\theta ) \\
\cos (3\theta ) = 4\cos ^3 (\theta ) - \cos (\theta ) \\
\end{array}\)

P.S. If one can type θ to get θ.

 

[tam14]

Thank you so much!

 

[soroban]

Hello, tam14!

I am trying to prove \(\displaystyle \cos3\theta\:=\:4\cos^3\theta\,-\,3\cos\theta\)

from Euler's formula: \(\displaystyle e^{i\theta}\:=\:\cos\theta\,+\,i\sin\theta\)

If we "cube" Euler's Formula:
\(\displaystyle \;\;\left(e^{i\theta}\right)^3 \;=\;(\cos\theta\,+\,i\sin\theta)^3\;=\;\cos^3\theta\,+\,3i\cos^2\theta\sin\theta \,- \,3\cos\theta\sin^2\theta\,- \,i\sin^2\theta\;\) [1]

But: \(\displaystyle \,\left(e^{i\theta}\right)^3\;=\;e^{3i\theta}\;=\;e^{i(3\theta)} \;=\;\cos3\theta\,+\,i\sin3\theta\;\) [2]


Since [1] and [2] are equal, we have:

\(\displaystyle \;\;\;\cos3\theta\,+\,i\sin3\theta \;= \;\cos3\theta\,+\,3i\cos^2\theta\sin\theta\,-\,3\cos\theta\sin^2\theta\,-\,i\sin^3\theta\)

or: \(\displaystyle \,\cos3\theta\,+\,i\sin3\theta \;= \;(\cos^3\theta\,-\,3\cos\theta\sin^2\theta)\,+\,i(3\cos^2\theta\sin\theta\,-\,\sin^3\theta)\)


Equating real and imaginary components, we have:

\(\displaystyle \cos3\theta\;=\;\cos^3\theta\,-\,3\cos\theta\sin^2\theta \;= \;\cos^3\theta\,-\,3\cos\theta(1\,-\,\cos^2\theta) \;=\;4\cos^3\theta\,-\,3\cos\theta\)

\(\displaystyle \sin3\theta\;=\;3\cos^2\theta\sin\theta\,-\,\sin^3\theta\;=\;3(1\,-\,\sin^2\theta)\sin\theta\,-\,\sin^3\theta \;= \;3\sin\theta\,-\,4\sin^3\theta\)


And we have the "Triple-Angle Identities" for both sine and cosine!


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If you desire the Triple-Angle Identity for tangent:

\(\displaystyle \tan3\theta \;= \;\L\frac{\sin3\theta}{\cos3\theta}\;= \;\frac{3\cos^2\theta\sin\theta\,-\,\sin^3\theta}{\cos^3\theta\,-\,3\cos\theta\sin^2\theta}\)


Divide top and bottom by \(\displaystyle cos^3\theta:\)

\(\displaystyle \tan3\theta\;=\;\L\frac{\frac{3\cos^2\theta\sin\theta}{\cos^3\theta}\,-\,\frac{\sin^3\theta}{\cos^3\theta}}{\frac{\cos^3\theta}{\cos^3\theta}\,-\,\frac{3\cos\theta\sin^2\theta}{\cos^3\theta}} \;= \;\frac{3\left(\frac{\sin\theta}{\cos\theta}\right)\,-\,\left(\frac{sin\theta}{\cos\theta}\right)^3}{1\,-\,3\left(\frac{\sin\theta}{\cos\theta}\right)^2} \;= \;\frac{3\tan\theta\,-\,\tan^3\theta}{1\,-\,3\tan^2\theta}\)