Re: Imaginary numbers
Hello, atomos!
Here's the first one . . .
Use the relationship: \(\displaystyle \,e^{i\theta}\:=\:\cos\theta\,+\,i\cdot\sin\theta\)
Express \(\displaystyle \cos5\theta\) in the terms of \(\displaystyle \cos\theta\)
That relationshp says: \(\displaystyle \,e^{i(5\theta)}\;=\;\cos(5\theta)\,+\,i\cdot\sin(5\theta)\;\)
[1]
We also have: \(\displaystyle \,e^{i5\theta}\;=\;\left(e^{i\theta}\right)^5\;=\;(\cos\theta\,+\,i\cdot\sin\theta)^5\)
\(\displaystyle \;\;= \;\cos^5\theta\,+\,5i\cdot\cos^4\theta\cdot\sin\theta\,-\,10\cdot\cos^3\theta\cdot\sin^2\theta\,-\,10i\cdot\cos^2\theta\cdot\sin^3\theta\,+\,5\cdot\cos\theta\cdot\sin^4\theta\,+\,i\cdot\sin^5\theta\)
\(\displaystyle \;\;=\;[\cos^5\theta\,-\,10\cdot\cos^3\theta\cdot\sin^2\theta\,+\,5\cdot\cos\theta\cdot\sin^4\theta]\,+\,i[5\cdot\cos^4\theta\cdot\sin\theta\,-\,10\cdot\cos^2\theta\cdot\sin^3\theta\,+\,\sin^5\theta]\;\)
[2]
Equate the real components of
[1] and
[2]:
\(\displaystyle \cos(5\theta) \;= \;\cos^5\theta\,-\,10\cdot\cos^3\theta\cdot\sin^2\theta\,+\,5\cdot\cos\theta\cdot\sin^4\theta\)
\(\displaystyle \;\;= \;\cos^5\theta\,-\,10\cdot\cos^3\theta\left(1\,-\,\cos^2\theta\right)\,+\,5\cdot\cos\theta\left(1\,-\,\cos^2\theta)^2\)
\(\displaystyle \;\;= \;\cos^5\theta\,-\,10\cdot\cos^3\theta\,+\,10\cdot\cos^5\theta\,+\,5\cdot\cos\theta\,-\,10\cdot\cos^3\theta\,+\,5\cdot\cos^4\theta\)
Therefore: \(\displaystyle \cos(5\theta) \;= \;16\cdot\cos^5\theta\,-\,20\cdot\cos^3\theta\,+\,5\cdot\cos\theta\)
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If we equate the imaginary components of
[1] and
[2], we have:
\(\displaystyle \sin(5\theta) \;= \;5\cdot\cos^4\theta\cdot\sin\theta\,-\,10\cdot\cos^2\theta\cdot\sin^3\theta\,+\,\sin^5\theta\)
\(\displaystyle \;\;=\;5(1\,-\,\sin^2\theta)^2\sin\theta \,- \,10\cdot(1\,-\,\sin^2\theta)\sin^3\theta\,+\,\sin^5\theta\)
\(\displaystyle \;\;=\;5\cdot\sin\theta\,-\,10\cdot\sin^3\theta\,+\,5\cdot\sin^5\theta\,-\,10\cdot\sin^3\theta\,+\,10\cdot\sin^5\theta \,+\,\sin^5\theta\)
Therefore: \(\displaystyle \,\sin(5\theta) \;= \;16\cdot\sin^5\theta\,-\,20\cdot\sin^3\theta\,+\,5\cdot\sin\theta\)
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