Question

State the amplitude and period of f(x) = 3cos9/5t

Answer 1

Answer:

The amplitude is 3 and the period is 10π/9.

Step-by-step explanation:

The standard cosine function is in the form:

$y=a\cos(b(x-c))+d$

Where |a| is the amplitude, 2π/b is the period, c is the phase shift, and d is the vertical shift.

We have the function:

$f(x)=\displaystyle 3\cos\frac{9}{5}t$

We can rewrite this as:

$\displaystyle f(x)=(3)\cos\left(\frac{9}{5}\left(t-0\right)\right)-(0)$

Therefore, a = 3, b = 9/5, c = 0, and d  = 0.

Hence, our amplitude is |3| = 3.

Our period will be:

$\displaystyle \text{Period}=\frac{2\pi}{9/5}=2\pi \cdot \frac{5}{9}=\frac{10\pi}{9}$