# Proof of inequality $|x \sin \alpha + y \cos \alpha| \leq \sqrt{x^2 + y^2}$ [duplicate]

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Just like in the title, I'm asking for any hints for proving (propably simple) inequality:

$$|x \sin \alpha + y \cos \alpha| \leq \sqrt{x^2 + y^2}$$

## marked as duplicate by Arnaud D., Carl Mummert, Thomas Shelby, Servaes, CesareoMar 2 at 0:50

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• Cauchy-Schwarz? – Lord Shark the Unknown Oct 8 '18 at 19:11
• Hint: Assuming $x^2+y^2 \neq 0$, we can divide both sides by $\sqrt{x^2+y^2}$. Then $\frac{x}{\sqrt{x^2+y^2}}$ can be thought of as $\sin \beta$, in which case $\frac{y}{\sqrt{x^2+y^2}}$ will be....? – Anurag A Oct 8 '18 at 19:12
• @AnuragA Of course, thanks! Should be obvious for me. – chandx Oct 8 '18 at 20:04

## 2 Answers

By C-S $$|x\sin\alpha+y\cos\alpha|\leq\sqrt{(\sin^2\alpha+\cos^2\alpha)(x^2+y^2)}=\sqrt{x^2+y^2}.$$

$$(x\cos t+y\sin t)^2+(x\sin t-y\cos t)^2=?$$