proof left as an exercise
06-09-2022, 12:35 AM
Post: #9
 Albert Chan Senior Member Posts: 2,006 Joined: Jul 2018
RE: proof left as an exercise
(06-08-2022 11:18 PM)Thomas Klemm Wrote:  We can use the triple angle formulae:

\begin{align} \sin(3 \theta) &= 3 \sin \theta - 4 \sin^{3} \theta \\ \cos(3 \theta) & = 4 \cos^{3}\theta - 3 \cos \theta \\ \end{align}

I noticed an easier way

sin(3θ)/sin(θ) = 4*cos(θ)^2 - 1
cos(3θ)/cos(θ) = 4*cos(θ)^2 - 3

sin(3θ)/sin(θ) = cos(3θ)/cos(θ) + 2

This is all is need for the proof:

2*cos(30°) / (1+4*sin(70°))
= 2*sin(60°) / (1+4*cos(20°))
= 2*sin(20°) * (cos(60°)/cos(20°) + 2) / (1+4*cos(20°))
= tan(20°) * (1+4*cos(20°)) / (1+4*cos(20°))
= tan(20°)
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 Messages In This Thread proof left as an exercise - Thomas Klemm - 06-06-2022, 11:41 PM RE: proof left as an exercise - Ángel Martin - 06-07-2022, 05:05 AM RE: proof left as an exercise - Thomas Klemm - 06-07-2022, 05:32 AM RE: proof left as an exercise - Albert Chan - 06-07-2022, 05:36 PM RE: proof left as an exercise - Albert Chan - 06-07-2022, 06:17 PM RE: proof left as an exercise - Albert Chan - 06-08-2022, 01:50 AM RE: proof left as an exercise - Albert Chan - 06-08-2022, 11:12 AM RE: proof left as an exercise - Thomas Klemm - 06-08-2022, 11:18 PM RE: proof left as an exercise - Albert Chan - 06-09-2022 12:35 AM RE: proof left as an exercise - Albert Chan - 07-01-2022, 07:51 PM RE: proof left as an exercise - Albert Chan - 07-02-2022, 11:44 PM

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