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I'd say I generally know how to do this, but I'm stuck on these two...

$1)\,\,\, y = 1/6\tan (x)$

I came up with $12\pi$, but apparently that's wrong.

$2) \,\,\,y = 5\csc (x - π/3)$

With this one I came up with $2π/5$, also wrong.

Help?

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What have you done to get those answers, perhaps it is only a slight mistake –  Jean-Sébastien Oct 24 '12 at 0:40
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2 Answers

up vote 2 down vote accepted

I guess it all goes back to knowing what $A$, $B$, $C$, and $D$ do to the trig expressions $A \tan(Bx \pm C) \pm D$ and $A \csc(Bx \pm C) \pm D$ .

If $A > 1$, then you have a "vertical stretch" of the graph of $\tan(x)$. If $0 < A < 1$, then you have a "vertical shrink" of the graph of the trig function. Then, notice that the period of the trig function is not affected by $A$ since $A$ doesn't change how long it takes to complete one cycle.

$+C$ will move the graph $C$ units to the left, and $-C$ will move the graph $C$ units to the right. Furthremore, $+D$ will move the graph $D$ units up, and $-D$ will move the graph $D$ units down. Since these values only shift the graph, they do not affect how long it takes to complete one cycle. Therefore, they don't change the period of the trig function.

However for $B$, this will affect the period of the graph. If $0 < B < 1$, then you have a "horizontal stretch" of the graph of the trig function, meaning that it will take longer to complete one cycle. If $B > 1$, then you have a "horizontal shrink" of the graph of the trig function, meaning that the graph will complete a cycle faster than normal.

So, the only time the period of the trig function changes is when you have a coeffiecient in front of the $x$.

Since $\tan(x)$ has a period of $\pi$, this means that $\frac{1}{6} \tan(x)$ would have a period of $\cdots$.

Since $\csc(x)$ has a period fo $2 \pi$, this means that $5 \csc(x - \frac{\pi}{3})$ would have a period of $\cdots$.

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$\bullet$ Note that if $f(x)$ has period $k,\;\;r\cdot f(x)$ also has period $k$. Why?

Set $g(x)=r\cdot f(x):\;\;f(x)=f(x+k)\iff r\cdot f(x)=r\cdot f(x+k)\iff g(x)=g(x+k)$

$\bullet$ Note that $f(x)$ and $f(x+s)$ both have the same period. The values are simply "shifted" by $s$.

So the first has the same period as $\tan$, and the second has the same period as $\csc$

Trick question!

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