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Given that $$5\cos \theta -12\sin \theta = 13$$ I'm trying to evaluate a general solution for this equation. It appears I'll be using vector product.

My equation is equivalent to

$$\langle (5,12), (\cos\theta, \sin\theta)\rangle = 13$$

which yields (by Cauch Schwarz Inequality) $$|\langle (5,12), (\cos\theta, \sin\theta)\rangle| \le \|(5,12)\|\|(\cos\theta, \sin\theta)\| = 13$$

This is where I'm stuck.

Regards

Blue
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Busi
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  • Do you know when the Cauchy-Schwarz inequality is actually an equality? – Arnaud D. Sep 03 '18 at 14:30
  • @ArnaudD. I truly do not. I'll be very glad if you can show. Even thought I googled it, there were not any useful results. That's what I'm actually missing. – Busi Sep 03 '18 at 14:30
  • Wait, you've pretty much asked this question before : https://math.stackexchange.com/questions/2903190/determining-the-general-solution-for-the-trigonometric-equation-5-cosx-12-si. If you have some trouble with some of the answers there, it's better to comment there. – Arnaud D. Sep 03 '18 at 14:33
  • @ArnaudD. Yes, that's because I did not get it properly. – Busi Sep 03 '18 at 14:33
  • So have you computed the length of each of the vectors? – Mark Bennet Sep 03 '18 at 14:44
  • I don't think there is any need for a new question. I've added a comment on the other question with a link to questions where the equality case of Cauchy-Schwarz is discussed. – Arnaud D. Sep 03 '18 at 14:44

2 Answers2

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Recall that given two vectors in $\mathbb{R^2}$ or $\mathbb{R^3}$ $u$ and $v$ by dot product we have

$$u\cdot v=|u||v|\cos \theta$$

and since $-1\le \cos \theta \le 1$ we have

$$-|u||v|\le u\cdot v\le |u||v|\iff |u\cdot v|\le |u||v|$$

Since $|\cos \theta|=1$ when $\theta=0, \pi$ the equality holds if and only if $u$ and $v$ are multiple vectors.

The result can be generalized for any dimension and it is known as Cauchy-Schwarz inequality.

user
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    I don't see how this answers the question. OP seems to know the Cauchy-Schwarz inequality. – Arnaud D. Sep 03 '18 at 14:41
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    @ArnaudD. Are you joking? Did you read the aswer of the OP to you own question?: Question: "Do you know when the Cauchy-Schwarz inequality is actually an equality?" - Answer: "I truly do not. I'll be very glad if you can show. Even thought I googled it, there were not any useful results. That's what I'm actually missing". – user Sep 03 '18 at 14:43
  • But your answer did not say anything about when the inequality is an equality, at the time I commented. – Arnaud D. Sep 03 '18 at 14:46
  • @ArnaudD. Please explain your comment or you have a very short memory or you are in bad faith! – user Sep 03 '18 at 14:46
  • @ArnaudD. I've answered after your question and after OP answer. I think you should apologize and pay attention to your behaviour before to accuse someone. – user Sep 03 '18 at 14:47
  • I don't think I should apologize for anything, but now that your answer at least gives the relevant idea, I've retracted my downvote. – Arnaud D. Sep 03 '18 at 14:53
  • @ArnaudD. I thought it was pretty clear from $u\cdot v=|u||v|\cos \theta$ that equality for $|u\cdot v|=|u||v||\cos \theta|$ holds for $cos \theta =1$. Anyway after your comment I've added some detail more on that even if you first comment wasn't referring to that point but to the inequality more in general. – user Sep 03 '18 at 14:53
  • @ArnaudD. You can downvote whenever you like but I can't accept bad faith accusations. If you don't want apologize we can close our discussion here. – user Sep 03 '18 at 14:55
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Hint. \begin{align} 5\cos \theta -12\sin \theta = 13 &\implies -12\sin \theta = 13-5\cos \theta \\ &\implies (-12\sin \theta )^2= (13-5\cos \theta)^2 \\ &\implies 144(1-\cos^2 \theta)= 169-130\cos \theta +\cos^2\theta \\ &\implies 144-144\cos^2 \theta= 169-130\cos \theta +\cos^2\theta \\ &\implies 145\cos^2 \theta-130\cos\theta+25=0 \end{align}

Elias Costa
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