# Math Help - Determining the rule for graphs of trig function

1. ## Determining the rule for graphs of trig function

A few question i am stuck on:

1) $y=Atan(nt)$ find A and n using the information provided
Asymptotes have equation $t=(2k+1)\frac{\pi}{6}$where k=>Z
when $t=\frac{\pi}{12},y=5$

2) $y=Asin(nt+e)$
Range = [-2,2]
Period=6
When t=1,y=1

I have found what A and n is however i don't know how to find e.

P.S

2. Originally Posted by Paymemoney
A few question i am stuck on:

1) $y=Atan(nt)$ find A and n using the information provided
Asymptotes have equation $t=(2k+1)\frac{\pi}{6}$where k=>Z
when $t=\frac{\pi}{12},y=5$

2) $y=Asin(nt+e)$
Range = [-2,2]
Period=6
When t=1,y=1

I have found what A and n is however i don't know how to find e.

P.S
2. You should know that for a sine function of the form

$y = a\sin{(bx)}$,

the amplitude is $a$ and the period is $\frac{2\pi}{b}$.

$y = A\sin{(nt + e)}$.

Since the range is $[-2, 2]$, that means the amplitude must be $2$.

Therefore $A = 2$.

The period is 6.

So $6 = \frac{2\pi}{n}$

$n = \frac{2\pi}{6}$

$n = \frac{\pi}{3}$.

So far you have

$y = 2\sin{\left(\frac{\pi t}{3} + e\right)}$.

You also know that when $t = 1, y= 1$.

So $1 = 2\sin{\left(\frac{\pi}{3} + e\right)}$

$\frac{1}{2} = \sin{\left(\frac{\pi}{3} + e\right)}$

$\frac{\pi}{6} = \frac{\pi}{3} + e$

$-\frac{\pi}{6} = e$.

Therefore $y = 2\sin{\left(\frac{\pi t}{3} - \frac{\pi}{6}\right)}$.