# Thread: Help with tangent function

1. ## Help with tangent function

Hi, here's my example from my math book that I don't understand.

Sketch the graph of y = -3 tan 2x.

By solving the equations

2x = - pi/2 and 2x = pi/2

x = -pi/2 x = pi/4

Here's the part that I have a problem understanding:

when x = - pi/8
- 3 tan 2x = 3

when x = 0
-3 tan 2x = 0

when x = pi/8
-3 tan 2x = -3

Can I do this calculation without looking at a graph. I can perform a simple x and y cordinate using this technique but when it comes to tangent problems, I can't same to apply the same principle.

2. Without looking at a graph?
Sure.

[Time will come, (when you are comfortable with Trigonometry already) when you don't have to look at graphs to analyze trig equations. Well, maybe seldom only.]

y = -3 tan 2x

----------------
when x = - pi/8,

y = -3 tan 2(-pi/8)
y = -3 tan -pi/4

angle -pi/4 is in the 4th quadrant.
Its reference angle is pi/4 (or 45 degrees). I assume you know by heart the tan(pi/4) or tan(45deg). It is 1.
Since -pi/4 is in the 4th quadrant, where tangent is negative (you know that?), then,

y = -3 *(-1)

---------------------
when x = 0

y = -3 tan 2*0
y = -3 tan 0

tan(0) is 0, so,

y = -3 * 0

-------------------------
when x = pi/8

y = -3 tan 2(pi/8)
y = -3 tan pi/4

Angle pi/4 (or 45 deg) is in the 1st quadrant.
You know that tan(pi/4) or tan(45deg) is 1.
So,

y = -3 *(1)