Therefore tan theta will equal zero whenever sin(\theta)=0

sin(\theta)=0 \newline \theta= ...,-\pi,0,\pi.....\newline \theta= k\pi

Where k is equal to any positive or negative integer and zero

It will be undefined when cos(\theta)=0 and an asymptote will occur there.

cos(\theta)=0 \newline \theta= ...,-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}.....\newline \theta= (2k+1)\frac{\pi}{2}

Where k is equal to any positive or negative integer and zero


The graph shows that as cos(\theta) \rightarrow 0, f(x) \rightarrow \infty^{+} if the angle is in the first and third quadrants, where tan is positive.

It will get closer to negative infinity if the angle is in the second and fourth quadrants, under the same conditions.


So let’s take a look at the first multiple choice question from the 2013 exam 2.


Step 1 calculate the period, which this question asks us to do and remember our formula for calculating the tan period is slightly different to when we calculuated sin and cos


So the answer is C.

From the period, as it is usually pi we can see that the function has been dialated by a factor of a \frac{1}{2\pi} from the y-axis.

Step 2: Find the y intercept let x=0

f(x)=-3tan(2\pi x)

Step 3 Find the x axis intercepts

Divide the tan theta x axis intercepts by \frac{1}{2\pi}


Where k is any whole integer for example:


Step 4: Find the x asymptotes

Apply the same dilation factor of \frac{1}{2\pi}


Where k is any whole integer number. Therefore:


Step 4: Dilation from the x-axis

It is also worth noting that the three at the front of the tan the function will stretch the function from the x-axis by a factor of three. Therefore at our values where tan normally equals one, halfway between the intercept and the asymptote will now equal three.

Step 5: Apply any reflections

Also due to the negative sign out the front the graph will reflect the graph in the x axis.