How do you graph y=3cosx?

Apr 21, 2018

See below:

Explanation:

We are going to graph it as a last step, but lets go through the different parameters of the sine and cosine functions. Im going to use radians when doing this by the way:

$f \left(x\right) = a \cos b \left(x + c\right) + d$

Parameter $a$ affects the amplitude of the function, normally Sine and Cosine have a maximum and minimum value of 1 and -1 respectively, but increasing or decreasing this parameter will alter that.

Parameter $b$ affects the period (but it is NOT the period directly)- instead this is how it affects the function:

Period= $\frac{2 \pi}{b}$

so a greater value of $b$ will decrease the period.

$c$ is the horizontal shift, so altering this value will shift the function either left or right.

$d$ is the principal axis that the function will revolve around ,normally this is the x-axis, $y = 0$, but increasing or decreasing the value of $d$ will alter that.

Now, as we can see the only thing affecting our function is the parameter $a$- which is equal to 3. This will effectively multiply all the values of the cosine function by 3, so now we can find some points to graph by plugging in some values:

$f \left(0\right) = 3 C o s \left(0\right) = 3 \times 1 = 3$

$f \left(\frac{\pi}{6}\right) = 3 C o s \left(\frac{\pi}{6}\right) = 3 \times \left(\frac{\sqrt{3}}{2}\right) = \frac{3 \sqrt{3}}{2}$

$f \left(\frac{\pi}{4}\right) = 3 C o s \left(\frac{\pi}{4}\right) = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$

$f \left(\frac{\pi}{2}\right) = 3 C o s \left(\frac{\pi}{2}\right) = 3 \times 0 = 0$

$f \left(\pi\right) = 3 C o s \left(\pi\right) = 3 \times - 1 = - 3$

(and then all the multiples of these numbers- but these should be sufficient for a graph)

Hence it will more or less look like this:

graph{3cosx [-0.277, 12.553, -3.05, 3.36]}