de moivres again (1 Viewer)

kractus · Jan 16, 2023

Screen Shot 2023-01-16 at 10.12.48 pm.png

I just let tan5x = sin5x/cos5x then use de moivres to get them in terms of their imaginary and real components but from there i cant seem to simplify it to RHS.

nathanzhou1234 · Jan 16, 2023

kractus said:
View attachment 37459
I just let tan5x = sin5x/cos5x then use de moivres to get them in terms of their imaginary and real components but from there i cant seem to simplify it to RHS.

Try dividing the top and bottom by cos until all u get is sin/cos terms to make into tan

Drongoski · Jan 17, 2023

$(cis \theta)^5 = cis 5 \theta \implies cos 5\theta + i sin 5\theta = (cos\theta +i sin \theta)^5 \equiv (c + is)^5 \\ \\ = c^5 +5ic^4s +10i^2c^3s^2 + 10 i^3c^2s^3 + 5i^4cs^4 + i^5s^5 \\ \\ = (c^5 -10 c^3 s^2 +5 cs^4) + i(5c^4 s - 10c^2s^3 + s^5)$

Equating reals and imaginaries:

$cos 5\theta = c^5 - 10c^3 s^2 + 5cs^4 $ and $ sin 5\theta = 5c^4 s - 10 c^2 s^3 + s^5 \\ \\ \therefore tan 5\theta = \frac {sin 5\theta}{cos 5 \theta} = \frac {5c^4s - 10 c^2 s^3 + s^5}{c^5 - 10 c^3s^2 + 5 cs^4} = \frac {\frac {5c^4s}{c^5} - \frac {10c^2s^3}{c^5} + \frac {s^5}{c^5}}{\frac {c^5}{c^5} - \frac {10c^3 s^2}{c^5} + \frac {4cs^4}{c^5}} = \frac {5t - 10t^3 + t^5}{1 - 10 t^2 + 5t^4} \\ \\ = \frac {5tan \theta - 10 tan^3 \theta + tan^5 \theta}{1 - 10 tan^2 \theta + 5 tan^4 \theta}$

QED

de moivres again (1 Viewer)

kractus

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nathanzhou1234

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Drongoski

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