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This question already has an answer here:

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

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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Cauchy-Schwarz? $\endgroup$ – Lord Shark the Unknown Oct 8 '18 at 19:11
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    $\begingroup$ 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....? $\endgroup$ – Anurag A Oct 8 '18 at 19:12
  • $\begingroup$ @AnuragA Of course, thanks! Should be obvious for me. $\endgroup$ – chandx Oct 8 '18 at 20:04
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By C-S $$|x\sin\alpha+y\cos\alpha|\leq\sqrt{(\sin^2\alpha+\cos^2\alpha)(x^2+y^2)}=\sqrt{x^2+y^2}.$$

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Use https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity,

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

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