# How do you calculate amplitude and period of y = -0.3 cos (2x)?

Apr 28, 2015

In this problem we deal with a type of function $F \left(x\right)$ that is called sinusoidal.
These functions have a period, that is there exists a real value $T$ such that $F \left(x + T\right) = F \left(x\right)$ for any $x$.
These functions also have a characteristic called an amplitude - the maximum deviation from zero occurred on any interval of at least a size of a period $T$.

If function $F \left(x\right)$ has a period $T$, then a function $F \left(k \cdot x\right)$ has a period of $\frac{T}{k}$.
This fact follows from a simple chain of equalities:
$F \left(k \cdot \left(x + \frac{T}{k}\right)\right) = F \left(k \cdot x + k \cdot \frac{T}{k}\right) =$
$= F \left(k \cdot x + T\right) = F \left(k \cdot x\right)$
(since $T$ is a period of $F \left(x\right)$ for any value of an argument, including $k \cdot x$).

We have proven that
$F \left(k \cdot \left(x + \frac{T}{k}\right)\right) = F \left(k \cdot x\right)$,
which means that $\frac{T}{k}$ is a period.

If a function $F \left(x\right)$ has an amplitude $A > 0$, then a function $C \cdot F \left(x\right)$ has an amplitude $| C | \cdot A$.
Indeed, let's consider that the maximum deviation of $F \left(x\right)$ on a segment $\left[0 , T\right]$ equals $A$ and it is reached at $x = {x}_{0}$, that is $F \left({x}_{0}\right) = | A |$.
Then $C \cdot F \left({x}_{0}\right) = C \cdot | A |$, from which follows that an amplitude of $C \cdot F \left(x\right)$ is, at least, equal to $| C \cdot A | = | C | \cdot A$ (as $A > 0$). It also cannot be greater that $| C | \cdot A$ since at that point of an $x$ the original function $F \left(x\right)$ would be greater than $A$.

Obviously, changes in period and altitude of a sinusoidal function are independent of each other and can be evaluated separately.

Using the above theoretical consideration, we see in our case for a function $y = - 0.3 \cos \left(2 x\right)$ that its period equals to $\frac{T}{2}$ (where $T$ is a period of a function $y = \cos \left(x\right)$) and its altitude equals to $| - 0.3 | \cdot A$ (where $A$ is an altitude of a function $y = \cos \left(2 x\right)$.

We know that the period of a function $y = \cos \left(x\right)$ equals to $2 \pi$. Therefore, the period of a function $y = \cos \left(2 x\right)$ equals to $\frac{2 \pi}{2} = \pi$.

The amplitude of a function $y = \cos \left(2 x\right)$ equals to $1$. Therefore, the amplitude of $y = - 0.3 \cos \left(2 x\right)$ equals to $| - 0.3 | \cdot 1 = 0.3$