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#### Alexmahone

##### Active member

- 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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