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Alexmahone
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- Jan 26, 2012
- 268
We know
$\displaystyle -\sqrt{A^2+B^2}\le A\sin x+B\cos x\le\sqrt{A^2+B^2}$
Replacing $\displaystyle x$ by $\displaystyle x+a$,
$\displaystyle -\sqrt{A^2+B^2}\le A\sin (x+a)+B\cos (x+a)\le\sqrt{A^2+B^2}$
$\displaystyle -\sqrt{A^2+B^2}\le A\sin x+B\cos x+A\cos a+B\sin a\le\sqrt{A^2+B^2}$
But $\displaystyle -\sqrt{A^2+B^2}\le A\sin x+B\cos x\le\sqrt{A^2+B^2}$.
So, $0\le A\cos a+B\sin a\le 0$
---------- Post added at 11:00 PM ---------- Previous post was at 10:52 PM ----------
I must be going crazy. I expanded $\displaystyle \sin (x+a)$ and $\displaystyle \cos(x+a)$ incorrectly. Never mind...
$\displaystyle -\sqrt{A^2+B^2}\le A\sin x+B\cos x\le\sqrt{A^2+B^2}$
Replacing $\displaystyle x$ by $\displaystyle x+a$,
$\displaystyle -\sqrt{A^2+B^2}\le A\sin (x+a)+B\cos (x+a)\le\sqrt{A^2+B^2}$
$\displaystyle -\sqrt{A^2+B^2}\le A\sin x+B\cos x+A\cos a+B\sin a\le\sqrt{A^2+B^2}$
But $\displaystyle -\sqrt{A^2+B^2}\le A\sin x+B\cos x\le\sqrt{A^2+B^2}$.
So, $0\le A\cos a+B\sin a\le 0$
---------- Post added at 11:00 PM ---------- Previous post was at 10:52 PM ----------
I must be going crazy. I expanded $\displaystyle \sin (x+a)$ and $\displaystyle \cos(x+a)$ incorrectly. Never mind...
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