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

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?

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 $x \cos \varphi + y \sin \varphi\le 1$.

It's enough to show that $|x\cos\varphi + y\sin\varphi| < 2$. The triangle inequality gives us $$|x \cos \varphi + y \sin \varphi| \leq |x|\cdot 1 + |y|\cdot1.$$

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

• The first line may not be true when $x$ and $\cos \varphi$ are negative. – Erick Wong Apr 8 '17 at 19:22
• @ErickWong Thank you for catching that, does it look okay now? – Andrew Apr 8 '17 at 19:37
• Yup, looks fine. Though you spelled AM-GM wrong :). – Erick Wong Apr 8 '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. – Andrew Apr 8 '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 8 '17 at 23:59

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 $$|x\cos\varphi+y\sin\varphi|\leq|x|\,|\cos\varphi|+|y|\,|\sin\varphi|\leq|x|+|y|<2.$$

• Cauchy-Schwarz gives an upper bound of $1$. This is much weaker, so "need" is a bit of an overstatement. – Erick Wong Apr 8 '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 8 '17 at 20:13