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

1 Answer
Apr 28, 2015

In this problem we deal with a type of function F(x) that is called sinusoidal.
These functions have a period, that is there exists a real value T such that F(x+T)=F(x) 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(x) has a period T, then a function F(k*x) has a period of T/k.
This fact follows from a simple chain of equalities:
F(k*(x+T/k))=F(k*x+k*T/k)=
=F(k*x+T)=F(k*x)
(since T is a period of F(x) for any value of an argument, including k*x).

We have proven that
F(k*(x+T/k))=F(k*x),
which means that T/k is a period.

If a function F(x) has an amplitude A>0, then a function C*F(x) has an amplitude |C|*A.
Indeed, let's consider that the maximum deviation of F(x) on a segment [0,T] equals A and it is reached at x=x_0, that is F(x_0)=|A|.
Then C*F(x_0)=C*|A|, from which follows that an amplitude of C*F(x) is, at least, equal to |C*A|=|C|*A (as A>0). It also cannot be greater that |C|*A since at that point of an x the original function F(x) 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=-0.3cos(2x) that its period equals to T/2 (where T is a period of a function y=cos(x)) and its altitude equals to |-0.3|*A (where A is an altitude of a function y=cos(2x).

We know that the period of a function y=cos(x) equals to 2pi. Therefore, the period of a function y=cos(2x) equals to (2pi)/2=pi.

The amplitude of a function y=cos(2x) equals to 1. Therefore, the amplitude of y=-0.3cos(2x) equals to |-0.3|*1=0.3