# How to do: Phase Shift of tan

• October 16th 2009, 09:43 PM
S2Krazy
How to do: Phase Shift of tan
I was wondering how do i do the phase shift for

http://www.webassign.net/www38/latex...81d6241275.gif

all i know is:

?? 3x ??
• October 17th 2009, 12:16 AM
Hello S2Krazy
Quote:

Originally Posted by S2Krazy
I was wondering how do i do the phase shift for

http://www.webassign.net/www38/latex...81d6241275.gif

all i know is:

?? 3x ??

The graph of $y = \tan(\theta+ \tfrac{\pi}{6})$ is the same as the graph of $y =\tan(\theta)$, shifted a distance $\tfrac{\pi}{6}$ to the left.

If you want to see why, just look at the zero values.
When $\theta = 0, y=\tan(\theta)=\tan(0) = 0$. So the graph 'starts' at $(0,0)$.
But when $\theta=0, y=\tan(\theta+\tfrac{\pi}{6})=\tan(\tfrac{\pi}{6})$. So this graph 'starts' at $(0,\tfrac{\pi}{6})$, and it's when $\theta = -\tfrac{\pi}{6}$ that we get the value $y = 0$.
Now if we multiply the independent variable by $3$, and replace $\theta$ with $3x$, we get the graphs of $y = \tan(3x)$ and $y = \tan(3x+\tfrac{\pi}{6})$. Now, in the second graph, we shall need to take $3x = -\tfrac{\pi}{6}$ to get our zero value of $y$. So that's $x = -\tfrac{\pi}{18}$. So the graph has now been shifted $\tfrac{\pi}{18}$ to the left.

Finally, note that multiplying by $5$, to give $y = 5\tan(3x+\tfrac{\pi}{6})$, makes no difference to the phase-shift: it merely 'stretches' the graph in a vertical direction either side of the x-axis by a factor of $5$.

• October 17th 2009, 10:23 AM
S2Krazy
ah icic cool thxs.

so for

http://www.webassign.net/www38/latex...d954a06db3.gif

the phase shift would be -4pi/5 correct?
• October 17th 2009, 10:54 AM
Quote:

Originally Posted by S2Krazy
ah icic cool thxs.

so for

http://www.webassign.net/www38/latex...d954a06db3.gif

the phase shift would be -4pi/5 correct?

Correct!