Determining the rule for graphs of trig function

**Paymemoney** · December 12th 2009, 09:46 PM

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

**Prove It** · December 12th 2009, 09:57 PM

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

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