Find the smallest number

**Eraser147** · November 1st 2012, 03:45 PM

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**MarkFL** · November 1st 2012, 03:47 PM

Hint: $\sin(\theta)=\sin(\theta+2k\pi)$ where $k\in\mathbb{Z}$ .

**Eraser147** · November 1st 2012, 03:50 PM

Nope. Textbook says this

**MarkFL** · November 1st 2012, 04:00 PM

Use the hint I gave in conjunction with the hint your book provides (which I assumed you already knew).

$\sin\left(\frac{\pi}{4} \right)=\frac{1}{\sqrt{2}}$

$\sin(\theta)=\sin(\theta +14\pi)$

**Eraser147** · November 1st 2012, 04:01 PM

Well, I'm not sure if this way works but it gave me the right answer. If $\sin(\theta)$ = Sqrt 2/2 then that means the radian must be at pi/4. So I came up with a formula that looks like this. (4n+1)pi over 4. So I just plugged in 14 into n and I got my answer.

**MarkFL** · November 1st 2012, 05:43 PM

Yes:

$\frac{1}{\sqrt{2}}=\sin\left(\frac{\pi}{4} \right)=\sin\left(\frac{\pi}{4}+14\pi \right)=\sin\left(\frac{(4\cdot14+1)\pi}{4} \right)=\sin\left(\frac{57\pi}{4} \right)$

**HallsofIvy** · November 1st 2012, 07:10 PM

There is NO such "smallest number" but there is an obvious smallest positive number.

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