proof left as an exercise
06-08-2022, 11:18 PM
Post: #8
 Thomas Klemm Senior Member Posts: 2,078 Joined: Dec 2013
RE: proof left as an exercise
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}

\begin{align} \sin(3 \theta) &= 3 \sin \theta - 4 \sin^{3} \theta \\ &= \sin \theta \, [3 - 4 \sin^{2} \theta] \\ &= \sin \theta \, [4 \cos^{2} \theta - 1] \\ \end{align}

$$2 \cos(3 \theta) - 1 = 8 \cos^{3}\theta - 6 \cos \theta - 1 \\$$

For $$\theta = 20^\circ$$ we get:

\begin{align} 2 \cos(3 \theta) - 1 &= \\ 2 \cos(3 \cdot 20^\circ) - 1 &= \\ 2 \cos(60^\circ) - 1 &= \\ 2 \cdot \tfrac{1}{2} - 1 &= 0 \\ \end{align}

And thus:

$$8 \cos^{3}(20^\circ) - 6 \cos(20^\circ) - 1 = 0$$

From this we conclude:

\begin{align} 8 \cos^{3}(20^\circ) - 2 \cos(20^\circ) &= 1 + 4 \cos(20^\circ) \\ 2 \cos(20^\circ) [4 \cos^{2}(20^\circ) - 1] &= 1 + 4 \cos(20^\circ) \\ \end{align}

Time to plug all that into the formula:

\begin{align} \frac{2\cos(30^{\circ})}{1+4\sin(70^{\circ})} &= \frac{2\sin(60^{\circ})}{1+4\cos(20^{\circ})} \\ \\ &= \frac{2\sin(3 \cdot 20^{\circ})}{1+4\cos(20^{\circ})} \\ \\ &= \frac{2 \sin(20^\circ) [4 \cos^{2}(20^\circ) - 1]}{2 \cos(20^\circ) [4 \cos^{2}(20^\circ) - 1]} \\ \\ &= \frac{\sin(20^\circ)}{\cos(20^\circ)} \\ \\ &= \tan(20^\circ) \end{align}

Or then we use the product to sum identity:

$$2 \cos \theta \cos \varphi = \cos(\theta + \varphi )+\cos(\theta - \varphi)$$

We use $$\cos(60^\circ) = \frac{1}{2}$$ and start with:

\begin{align} \cos(10^\circ) &= 2 \cos(60^\circ) \cos(10^\circ) \\ &= \cos(70^\circ) + \cos(50^\circ) \\ \\ \cos(50^\circ) + \cos(10^\circ) &= \cos(70^\circ) + 2 \cos(50^\circ) \\ 2 \cos(30^\circ) \cos(20^\circ) &= \sin(20^\circ) + 2 \sin(40^\circ) \\ &= \sin(20^\circ) + 4 \sin(20^\circ) \cos(20^\circ) \\ &= \sin(20^\circ)[1 + 4 \cos(20^\circ)] \\ \end{align}

\begin{align} \frac{2 \cos(30^\circ)}{1 + 4 \cos(20^\circ)} = \frac{\sin(20^\circ)}{\cos(20^\circ)} = \tan(20^\circ) \end{align}

Or then:

\begin{align} \frac{2 \cos(30^\circ)}{1 + 4 \sin(70^\circ)} = \tan(20^\circ) \end{align}
 « Next Oldest | Next Newest »

 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

User(s) browsing this thread: 1 Guest(s)