# How do you find the amplitude, period, phase shift given y=3sinpix-5cospix?

Oct 17, 2016

Amplitude is $\sqrt{34}$, period is $2$ and phase shift is $\frac{1}{\pi} {\tan}^{- 1} \left(- \frac{5}{3}\right)$.

#### Explanation:

We have $y = 3 \sin \pi x - 5 \cos \pi x$ (Note ${3}^{2} + {5}^{2} = 34$)

= $\sqrt{34} \left(\frac{3}{\sqrt{34}} \sin \pi x - \frac{5}{\sqrt{34}} \cos \pi x\right)$

= $\sqrt{34} \left(\sin \pi x \cos \alpha - \cos \pi x \sin \alpha\right)$

= $\sqrt{34} \sin \left(\pi x - \alpha\right)$,

where, as $\sin \alpha = \frac{5}{\sqrt{34}}$ and $\cos \alpha = \frac{3}{\sqrt{34}}$,

$\alpha = {\tan}^{- 1} \left(- \frac{5}{3}\right)$

Now, as in $y = r \sin \left(p x + q\right)$

amplitude is $r$, period is $\frac{2 \pi}{p}$ and phase shift is $- \frac{q}{p}$.

In $y = 3 \sin \pi x - 5 \cos \pi x = \sqrt{34} \sin \left(\pi x - \alpha\right)$

amplitude is $\sqrt{34}$, period is $\frac{2 \pi}{\pi} = 2$ and phase shift is $\frac{\alpha}{\pi} = \frac{1}{\pi} {\tan}^{- 1} \left(- \frac{5}{3}\right)$.