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So I have a couple of quick little questions about trig functions. First, if you have a equation that looks something like $\sin x - 5 \cos x=0$, should you read it as $(\sin x)-(5 \cos x)$, or $(\sin (x)-5)(\cos x)$?

I feel like this is a pretty basic order of operations thing; I'm just having a little trouble applying it to trig functions.

Also, if you have, for example, $y = \sin \frac{\pi}{4}(x + 1)$, the phase shift remains $1$, an the periodizer $\frac{\pi}{4}$ -- the $\frac{\pi}{4}$ doesn't distribute.

But if I get $y = \sin (\frac{\pi}{4}x + \frac{\pi}{4})$, for example, do I take out the $\frac{\pi}{4}$ to get $y = \sin \frac{\pi}{4}(x + 1)$?

Thanks!

evamvid

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In general, functions generally follow the format $f(x)$ and any term such as $f(x)a$ is read $f(x)$ multiplied by $a$.

So in this case, only the number right after the trigonometric function should be taken as part of the function, unless it's in brackets. However, not everyone follows this formatting, so the question itself should give indication to what is intended.

Yes, you can distribute out the $\frac{\pi}{4}$, but make sure you're clear that the whole expression is still part of the trigonometric function. For your first question, it is $(sin(x)) - (5cos(x)) = 0$

Edit:

A function $f(x) = a + bsin(m(x - s))$ has amplitude $|b|$, period $\frac{2\pi}{m}$, phase shift $s$, and vertical shift $a$.

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  • $\begingroup$ So (for the second one), do I choose whether or not to distribute the $\frac{\pi}{4}$? If so, what is the phase shift; $1$, or $\frac{\pi}{4}$? $\endgroup$ – evamvid Jan 31 '14 at 14:42
  • $\begingroup$ You would factor out the $pi/4$ to find the phase shift. $\endgroup$ – Chantry Cargill Jan 31 '14 at 14:45
  • $\begingroup$ A function f(x) = a + bsin(m(x - s)) has amplitude |b|, period 2pi/m, phase shift s, and vertical shift a. $\endgroup$ – Chantry Cargill Jan 31 '14 at 14:47
  • $\begingroup$ So just to confirm, you take out the $\frac{\pi}{4}$ to get the phase shift. Right? $\endgroup$ – evamvid Jan 31 '14 at 14:56
  • $\begingroup$ Yes. So in this case, it would be 1. $\endgroup$ – Chantry Cargill Jan 31 '14 at 14:58

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