# How do you find the amplitude and period of f(x) = - 8 sin(5*x + pi) ?

Jul 25, 2015

Amplitude = $8$
Period = $\frac{2 \pi}{5}$

#### Explanation:

$f \left(x\right) = - 8 \sin \left(5 x + \pi\right)$
$\implies f \left(x\right) = 8 \left(- \sin \left(5 x + \pi\right)\right)$
$\implies f \left(x\right) = 8 \sin \left(5 x + \cancel{\pi} - \cancel{\pi}\right)$ (Since $- \sin \left(x\right) = \sin \left(x - \pi\right)$
$\implies f \left(x\right) = 8 \sin \left(5 x\right)$

Since $| \sin \left(x\right) | \le 1 \forall x$, the amplitude of $f \left(x\right)$ is 8.

Now, for the period, notice the following:
 f( x + (2pi)/5) = 8 sin( 5 ( x + (2pi)/5 ) ) = 8 sin( 5x + 2pi ) ) = 8 sin( 5x ) = f(x)

Thus the period of $f \left(x\right)$ is $\frac{2 \pi}{5}$.

Note: In general, for any sinusoidal function of the form $\sin \left(\nu x\right)$ or $\cos \left(\nu x\right)$, the period is $\frac{2 \pi}{\nu}$. Here $\nu$ is the frequency multiplication factor. In this particular case, $\nu = 5$.