Graph y = -tan(π/2). What are its vertical asymptotes?

How would I graph the function?

Thanks!

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- Jun 1st 2009, 06:04 PMlive_laugh_luv27Graph of trig function
Graph y = -tan(π/2). What are its vertical asymptotes?

How would I graph the function?

Thanks! - Jun 1st 2009, 06:59 PMartvandalay11
$\displaystyle y=-\tan\left(\frac{n}{2}\right)=\ -\frac{\sin\left(\frac{n}{2}\right)}{\cos\left(\fra c{n}{2}\right)}$ and if you know the half angle formulas, which state

$\displaystyle \sin\left(\frac{n}{2}\right)=\pm\sqrt{\frac{1-\cos(n)}{2}}$

$\displaystyle \cos\left(\frac{n}{2}\right)=\pm\sqrt{\frac{1+\cos (n)}{2}}$

Then $\displaystyle -\frac{\sin\left(\frac{n}{2}\right)}{\cos\left(\fra c{n}{2}\right)}=\pm\sqrt{\frac{1-cos(n)}{1+cos(n)}}=\pm\sqrt{\frac{1-cos(n)}{1+cos(n)}*\frac{1-cos(n)}{1-cos(n)}}=\pm\sqrt{\frac{(1-cos(n))^2}{1-cos^2(n)}}$

$\displaystyle =\pm\sqrt{\frac{(1-cos(n))^2}{sin^2(n)}}=\frac{1-cos(n)}{sin(n)}$

Now finding your asymptotes should be easy (sin(n)=0) and graphing this by hand, well I would graph the numerator and then the denominator, and then divide the corresponding y values point by point for an accurate sketch

as theemptyset has just pointed out, seems as though i may have misinterpretted that pi for an n, making this wrong and yeah... sorry for that - Jun 1st 2009, 07:01 PMTheEmptySet
I am going to assume that you mean

$\displaystyle y=-\tan\left( \frac{\pi}{2} {\color{red}x}\right)$

The negative flips the graph of the tangent, and the $\displaystyle \frac{\pi}{2}$ compresses the peroid of the function to $\displaystyle (-1,1)$

So the graph is

Attachment 11721