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

Question

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

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

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 08 '18 at 19:12

score 3 · Accepted Answer · answered Oct 08 '18 at 19:25

3

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

answered Oct 08 '18 at 19:25

Michael Rozenberg

score 0 · Answer 2 · answered Oct 08 '18 at 19:37

0

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

answered Oct 08 '18 at 19:37

lab bhattacharjee

2 Answers2