# Determining the rule for graphs of trig function

• Dec 12th 2009, 09:46 PM
Paymemoney
Determining the rule for graphs of trig function
A few question i am stuck on:

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

2)$\displaystyle y=Asin(nt+e)$
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
• Dec 12th 2009, 09:57 PM
Prove It
Quote:

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

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

2)$\displaystyle y=Asin(nt+e)$
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

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

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

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

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

Therefore $\displaystyle A = 2$.

The period is 6.

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

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

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

So far you have

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

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

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

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

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

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

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