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I have a question on this test review problem (that will help us on a test), and I have no clue what it's asking. We're learning trigonometry, (Analytic Trigonometry), like about the unit circle, inverse trig functions, etc... And I encounter this problem on test review:

$$ 2\cos\left(2\theta\right) = \sqrt{3}$$

I know the answers are: $\left\{\frac{\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{23\pi}{12} \right\} $

But I want to know how to find it. So I get ready for test. Thank you.

EDIT

After some thinking i got some of it but not All!! So what i did was:

$ 2\Theta = \cos^{-1} (\frac{\sqrt3}{2} )$

$2\Theta = \frac {\pi}{6}$

$\Theta = \frac{\pi}{12} + 2\pi n$

So now i just add 2pi but i have to remember $0 \le \Theta \lt 2\pi$

I get:

$\left\{\frac{\pi}{12}, \frac{13\pi}{12} \right\} $ But that obviously isnt the anwser.. Please help!

N. F. Taussig
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amanuel2
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    $\frac{\sqrt{3}}{2}$ should be recognisable as coming from half an equilateral triangle, so is the cosine of $\frac{\pi}{6}$ and various related angles which are possible values of $2\theta$. If $0 \le \theta \lt 2\pi$, you want to find the related angles such that $0 \le 2\theta \lt 4\pi$ – Henry Apr 29 '16 at 22:30
  • Huh @Henry ? You sound like my math teacher now :P – amanuel2 Apr 29 '16 at 22:41
  • Here's an answer of mine to a similar question, that uses the technique @Henry is describing. – pjs36 Apr 29 '16 at 22:47
  • This question does NOT deserve upvotes. The quality of MSE is going down.. – N.S.JOHN May 01 '16 at 06:05
  • You forgot that $\cos(x)=\cos(-x)$. –  May 17 '16 at 14:56

4 Answers4

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This writes as $\;\cos2\theta=\dfrac{\sqrt 3}2=\cos\dfrac\pi6$. Now you have to know that

  • $\cos x=\cos\alpha\iff\theta\equiv\pm\alpha\pmod2\pi$, and similarly:
  • $\sin x=\sin \alpha\iff\begin{cases}\theta\equiv\alpha\pmod{2\pi},\\\theta\equiv\pi-\alpha\pmod{2\pi}, \end{cases}$
  • $\tan x=\tan\alpha\iff x\equiv\alpha\pmod\pi.$

So here, you have $$2\theta\equiv \pm\frac\pi6\pmod{2\pi}\iff \theta\equiv\frac\pi{12}\pmod{\pi}$$ If you want the solutions in $[0,2\pi]$, you get $$\theta =\frac\pi{12},-\frac\pi{12}+\pi=\frac{11\pi}{12},\frac\pi{12}+\pi=\frac{13\pi}{12},-\frac\pi{12}+2\pi=\frac{23\pi}{12}.$$

Bernard
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  • We never, learned about the equation you use... mod $$2 \pi$$ ?? – amanuel2 Apr 29 '16 at 23:02
  • @AmanuelBogale Essentially that is saying $$2\theta = 2n\pi \pm \frac\pi 6$$ for integers $n$. – peterwhy Apr 29 '16 at 23:08
  • That's congruence, as in arithmetic, a very handy way to express solutions. Explicitly, $\theta\equiv\dfrac\pi{12}\pmod{\pi}$ means there exists an integer $k$ such that $\theta =\dfrac\pi{12}+k\pi$. So all we have to do is to draw up the list of those elements which are in the interval $[0,2\pi]$. – Bernard Apr 29 '16 at 23:08
  • ?? I am soo confused now @peterwhy , Bernard.. – amanuel2 Apr 29 '16 at 23:23
  • @Bernard i understand some part of it now.. But i didnt understand the rest. I will edit my question. Please Explain me what i am doing wrong when i edit the question so i can get a good grade on test. Thanks a bunch! – amanuel2 Apr 30 '16 at 16:38
  • Ok i edited question @peterwhy and Bernard. Please Help! I got some of it but not all! – amanuel2 Apr 30 '16 at 16:50
  • Ok guys... Forget my last comment. I Understand now! Thanks a bunch for this! – amanuel2 Apr 30 '16 at 18:31
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I would reform the equation $2\cos 2\theta = \sqrt{3}$ into $$\cos 2\theta \ =\frac{\sqrt 3}{2}$$

For now, I wouldn't worry about the $2\theta$ because if $\theta$ is an angle, then so is $2\theta$, so my intermediary strategy is to find $2\theta$. Later I can always divide that number by 2.

So lets fund $2\theta$. I'm looking for a triangle that has an adjacent side of $\sqrt 3$ and a hypotenuse of $2$. If I take an equilateral triangle, and split it in half, I will have such a triangle. enter image description here

So I know that that the angle that I'm looking for is $30^\circ$, or $\frac{\pi}{6}$ rads. In other words, $2\theta = \frac{\pi}{6}$, solving for $\theta$ we have $\theta = \frac{\pi}{12}$.

Now that you have your reference angle all figured out, try to figure out what quadrants will give you a positive cosine. I think that you will find that the angles are $\theta = (2\pi)n\pm\frac{\pi}{12}$

John Joy
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Perhaps the included picture will help you understand the solutions.enter image description here. The figure shows $\cos \theta$ in the range $-2 \pi \le \theta \le 2 \pi$. The blue line is the value $\cos \frac{\pi}{12}$ and you can see that it cuts the curve at four values in the request range. It's difficult to make out the required solutions (but you know them anyway), but it may help you understand whey there are four answers.

jim
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we have $\cos \frac{\pi}{6} = \frac{\sqrt{3}}2$. Hence your equation reads $$ \cos 2\theta = \cos \frac{\pi}{6}$$ then $$2\theta = \frac{\pi}{6} + 2k\pi \ \ \mbox{ or } 2\theta = -\frac{\pi}{6} + 2k\pi \ \ \ \ \ \ \ (k\in\mathbb{Z})$$ that is $$\theta = \frac{\pi}{12} + k\pi \ \ \mbox{ or } \theta = -\frac{\pi}{12} + k\pi \ \ \ \ \ \ \ (k\in\mathbb{Z})$$

It is clear that there are an infinite number of solution not only 4. (But if you precise an appropriate interval for solutions, you can make it to have those solutions in your question )

Aymane Fihadi
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