Question

Right now dealing with transformations of sine and cosine functions. Here's the equation I'm stumped on:
F (x)=3 cos (4 pi x - (pi/2)) - 2

I'm trying to put this into the form:
A cos [B (x - C)] + D
But I can't figure out how to factor the
(4 pi x - (pi/2)) into the form [B(x - C)].
I'm confused because if B is 4pi, and the period is (2pi/B), wouldn't I end up with a pi-less period? I haven't encountered that yet.
Also, if the phase shift is C/B, wouldn't I also end up with a pi-less phase shift?

Answer 1

Answer on Question #60763 – Math – Algebra

Question

Right now dealing with transformations of sine and cosine functions. Here's the equation I'm stumped on:

$$F(x) = 3 \cos \left(4pi x - (pi/2)\right) - 2$$

I'm trying to put this into the form:

$$A \cos [B(x - C)] + D$$

But I can't figure out how to factor the

$$(4pi x - (pi/2)) \text{ into the form } [B(x - C)].$$

I'm confused because if B is 4pi, and the period is (2pi/B), wouldn't I end up with a pi-less period? I haven't encountered that yet.

Also, if the phase shift is C/B, wouldn't I also end up with a pi-less phase shift?

Solution

We have got an expression

$$F(x) = 3 \cos \left(4\pi x - \frac{\pi}{2}\right) - 2.$$

Let's rewrite it in other form.

First of all,

$$\cos \left(4\pi x - \frac{\pi}{2}\right) = \cos \left(-\left(\frac{\pi}{2} - 4\pi x\right)\right) = \cos \left(\frac{\pi}{2} - 4\pi x\right) = \sin(4\pi x).$$

So, we have

$$F(x) = 3 \cos \left(4\pi x - \frac{\pi}{2}\right) - 2 = 3 \sin(4\pi x) - 2.$$

The graph of $F(x)$ is given below.

The period of $\sin(4\pi x)$, $3 \sin(4\pi x)$, $3 \sin(4\pi x) - 2$ is $\frac{2\pi}{4\pi} = \frac{1}{2}$, hence the period of $\cos\left(4\pi x - \frac{\pi}{2}\right)$.

$3 \cos \left(4\pi x - \frac{\pi}{2}\right) - 2$ will also be $\frac{2\pi}{4\pi} = \frac{1}{2}$. You will end up with a pi-less period. Also, if the phase shift is C/B, you will end up with a pi-less phase shift and it's Ok.

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