Showing that $x \cos \varphi + y \sin \varphi < 2$

Question

Let $S^1$ be the unit sphere and $D^2$ the unit disk in $\mathbb{R}^2$. Let $(\varphi, x, y) \in S^1 \times D^2$. I can't show that $$x \cos \varphi + y \sin \varphi < 2.$$ Can someone give me a hint?

What's the largest $x$ can be? What's the largest $\cos \varphi$ can be? — Erick Wong, Apr 08 '17 at 18:16
Possible duplicate of Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$ — Arnaud D., Feb 28 '19 at 15:14

Martin Argerami · Answer 1 · 2017-04-08T20:16:01.347

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You can use the Cauchy-Schwarz inequality. That is, $$ |x\,\cos\varphi+y\,\sin\varphi|\leq\sqrt{x^2+y^2}\,\sqrt{\cos^2\varphi+\sin^2\varphi}=\sqrt{x^2+y^2}\leq1<2. $$

A rather simpler argument is to notice that $x,y$ cannot be simultaneously $1$ (and neither $\cos\varphi$ and $\sin\varphi$, of course), so $$ |x\cos\varphi+y\sin\varphi|\leq|x|\,|\cos\varphi|+|y|\,|\sin\varphi|\leq|x|+|y|<2. $$

edited Apr 08 '17 at 20:16

answered Apr 08 '17 at 18:11

Martin Argerami

205,756

Cauchy-Schwarz gives an upper bound of $1$. This is much weaker, so "need" is a bit of an overstatement. – Erick Wong Apr 08 '17 at 18:17
Good point. I guess I'm programmed to see CS, so the triangle inequality is my second option, not my first. In any case, to get the strict inequality at 2, the triangle inequality is not enough. – Martin Argerami Apr 08 '17 at 20:13

Eman Yalpsid · Answer 2 · 2017-04-08T19:36:51.070

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Secondly, from the AM-QM inequality we get that $$|x| + |y| \leq 2 \sqrt{\frac{|x|^2+|y|^2}{2}} = 2 \frac{1}{\sqrt2} < 2.$$

edited Apr 08 '17 at 19:36

answered Apr 08 '17 at 18:42

Eman Yalpsid

3,044

The first line may not be true when $x$ and $\cos \varphi$ are negative. – Erick Wong Apr 08 '17 at 19:22
@ErickWong Thank you for catching that, does it look okay now? – Eman Yalpsid Apr 08 '17 at 19:37
Yup, looks fine. Though you spelled AM-GM wrong :). – Erick Wong Apr 08 '17 at 19:45
@ErickWong Thanks for checking it! By AM-QM I meant the relationship between the arithmetic and the quadratic mean... I wasn't sure how to call it. – Eman Yalpsid Apr 08 '17 at 19:57
Ah, I see yes you are using the quadratic mean. I have never heard of it referred to as the AM-QM inequality, but it's a reasonable name so I take back the earlier comment. – Erick Wong Apr 08 '17 at 23:59
@ErickWong Great, thank you again! :) – Eman Yalpsid Apr 09 '17 at 10:19

score 1 · Accepted Answer · answered Apr 08 '17 at 19:25

Each of the quantities $x,y,\cos \varphi, \sin \varphi$ is at most $1$ in magnitude, so $x \cos \varphi + y \sin \varphi \le 1\cdot 1 + 1 \cdot 1 = 2$.

The only way to achieve equality is if $x$ and $y$ both have magnitude $1$, which cannot happen, so the inequality is actually strict: $x \cos \varphi + y \sin \varphi < 2$.

In fact, as Martin Argerami points out, one can use Cauchy-Schwarz to obtain the tight bound $x \cos \varphi + y \sin \varphi\le 1$.

Showing that $x \cos \varphi + y \sin \varphi < 2$

3 Answers3