Solutions to equation involving trigonometry and geometric series

Question

$$\cos^2 x + \cos ^3 x +\dots = 1+ \cos x$$

I want to find values of $x$ between $0$ and $ 180$ degrees for which the above equation holds true.

Attempt at a solution: left side is a converging geometric progression, for which $a_1$ is $\cos^2 x$ and $q = \cos x$. Plugging into the known formula for such series yields $\cos^2 x = 1$, which yields $45 $ and $135$ degrees as solutions.

Is this okay?

Yes, indeed! I'll post a full solution too. – Ahaan S. Rungta Nov 01 '13 at 17:25 — Ahaan S. Rungta, Nov 01 '13 at 17:25
By the way, ^\circ gives you the degree symbol. – Stefan Smith Nov 01 '13 at 23:48 — Stefan Smith, Nov 01 '13 at 23:48

score 3 · Accepted Answer · answered Nov 01 '13 at 17:27

3

As mentioned, the LHS is a geometric series, with first term $ \cos^2 x $ and common ratio $ \cos x $, so it evaluates to $ \dfrac {\cos x}{1 - \cos x} $ and our equation becomes $$ \dfrac {\cos^2 x}{1-\cos x} = 1 + \cos x \implies \cos^2 x = 1 - \cos^2 x \implies \cos^2 x = \dfrac {1}{2}. $$Finally, take the square root of both sides and we have $ \cos x = \pm \dfrac {\sqrt{2}}{2} $ and our answers are $ \boxed {\dfrac{\pi}{4}, \dfrac{3\pi}{4}} $.

Seems I've been beaten by a couple of minutes by ncmathsadist and a few seconds by Oliver!

answered Nov 01 '13 at 17:27

Ahaan S. Rungta

7,706

Nonetheless awesome job mate! – Bak1139 Nov 01 '13 at 17:30
Thanks! Accept please, no? :) – Ahaan S. Rungta Nov 01 '13 at 17:30
Sorry for the double-comment but it seems my answer was accepted; thanks, Bak! – Ahaan S. Rungta Nov 01 '13 at 17:38
@Bak1139, if we are allowed to use this(http://mathworld.wolfram.com/Double-AngleFormulas.html), we have $\cos2x=2\cos^2x-1=0\implies 2x=(2n+1)\frac\pi2$ where $n$ is an integer – lab bhattacharjee Jan 24 '14 at 18:40

score 2 · Answer 2 · answered Nov 01 '13 at 17:23

2

Use a geometric sum. the left-hand side evaluates to $${\cos^2(x) \over 1 - \cos(x)},$$ provided $\cos(x) \not=\pm 1$. Can you do the rest?

answered Nov 01 '13 at 17:23

ncmathsadist

49,383

I performed that...my final line was cos^2 x = 0.5 – Bak1139 Nov 01 '13 at 17:26
Yep, it's correct. See Oliver's solution or my solution. – Ahaan S. Rungta Nov 01 '13 at 17:28

score 2 · Answer 3 · answered Nov 01 '13 at 17:26

2

On the LHS you have a geometric series.

$$\frac{\cos^2{x}}{1-\cos{x}} = 1 + \cos{x}$$

$$\cos^2{x} = 1 - \cos^2{x}$$ $$\cos{x} =\pm\frac{1}{\sqrt{2}}$$ $$x = \frac{\pi}{4}, \frac{3\pi}{4}$$

answered Nov 01 '13 at 17:26

user85798

1

its funny...I got that too...for some reason the official solutions state 45 and 150 degrees instead. – Bak1139 Nov 01 '13 at 17:27
Obviously a mistake. The two solutions are $ 45^\circ $ and $ 135^\circ $. You should accept our answers. – Ahaan S. Rungta Nov 01 '13 at 17:28

Solutions to equation involving trigonometry and geometric series

3 Answers3