# Math Help - Help with evaluating sine

1. ## Help with evaluating sine

I have the function
$f:[-\pi,\pi] \rightarrow R, f(x) = \sin 3(x+ \frac{\pi}{4})$

and I need to find the endpoints by subbing in $-\pi$ and $\pi$

$f(-\pi) = \sin 3(-\pi + \frac{\pi}{4})$
$= \sin 3(\pi+\frac{\pi}{4})$
$= \sin (3\pi+\frac{3\pi}{4})$
$= \sin (2\pi+\frac{7\pi}{4})$
$= \sin (\frac{7\pi}{4})$
Is this right so far? If so, what do I do next?

Thanks.

2. Originally Posted by user_5
I have the function
$f:[-\pi,\pi] \rightarrow R, f(x) = \sin 3(x+ \frac{\pi}{4})$

and I need to find the endpoints by subbing in $-\pi$ and $\pi$

$f(-\pi) = \sin 3(-\pi + \frac{\pi}{4})$
$= \sin 3(\pi-\frac{\pi}{4})$

Is this right so far? If so, what do I do next?

Thanks.
$\sin\left[3\left(\pi - \frac{\pi}{4}\right)\right]$

$\sin\left[3\left(\frac{3\pi}{4}\right)\right]$

$\sin\left(\frac{9\pi}{4}\right)$

fyi, $\frac{9\pi}{4}$ is coterminal with $\frac{\pi}{4}$

you should find the sine value on your unit circle.

3. $\sin \frac{\pi}{4} = \frac{1}{\sqrt2}$

The answer I have in my book for $f(-\pi) = -\frac{1}{\sqrt2}$

4. Originally Posted by user_5
$\sin \frac{\pi}{4} = \frac{1}{\sqrt2}$

this is $\textcolor{red}{f(\pi)}$
$f(-\pi) = \sin\left[3\left(-\pi + \frac{\pi}{4}\right)\right]$

$f(-\pi) = \sin\left[3\left(-\frac{3\pi}{4}\right)\right]$

$f(-\pi) = \sin\left(-\frac{9\pi}{4}\right) = -\sin\left(\frac{9\pi}{4}\right)$

5. Originally Posted by user_5
$\sin \frac{\pi}{4} = \frac{1}{\sqrt2}$

The answer I have in my book for $f(-\pi) = -\frac{1}{\sqrt2}$

That's wrong, the sine of $\frac{\pi}{4}$ (a 45 degree angle) is $\frac{\sqrt2}{2}$ all you have to do in that problem is figure out if that answer is positive or negative. The way to do that is first evaluate both of your functions with their inputs:

$f(-\pi) = \sin 3(-\pi + \frac{\pi}{4})$

$f(-\pi) = \sin (-3\pi + \frac{3\pi}{4})$

$f(-\pi) = \sin (-\frac{12\pi}{4} + \frac{3\pi}{4})$

$f(-\pi) = \sin (-\frac{9\pi}{4})$

Now, $-\frac{9\pi}{4}$ is coterminal with $\frac{\pi}{4}$ meaning when you go around the unit circle 9 times after $\frac{\pi}{4}$ you end up at the same place. Since sin is the y-axis of the unit circle and $-\frac{9\pi}{4}$ ends up being in the 4th quadrant of the graph, your answer will be negative. Therefore it should be:

$-\frac{\sqrt2}{2}$

EDIT: Well now that I think about it: $\frac{1}{\sqrt2} = \frac{\sqrt2}{2}$ so I guess your book is right :P

6. $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$