Trig Function Help

Atav20 · #1 $Old$ November 17th, 2008, 06:25 PM

Problem: Let f (x) = -cos(5x+(pie/3)). Determine the following:
a). the period of (fx)
b). the amplitude of f(x)
c). teh phase shift of f(x)

I think the answers are: a. 2pie/5 b. 1 c. -pie/15

Can someone plese check these answers and try to explain to me in a simple way how to find them. I really confuse myself.

Thanks

skeeter · #2 $Old$ November 17th, 2008, 06:44 PM

your solutions are correct

for $y = A\cos[B(x+C)] + D$

$|A|$ is the amplitude

$\frac{2\pi}{|B|}$ is the period

$C$ is the phase shift ... (+) left , (-) right

$D$ is the vertical shift ... (+) up, (-) down

btw ... for future reference, the Greek letter $\pi$ is spelled "pi".

this is a piece of "pie".

Atav20 · #3 $Old$ November 17th, 2008, 07:03 PM

Thanks for the grammar tip!

Atav20 · #4 $Old$ November 17th, 2008, 07:09 PM

Problem: For which values of x in the interval {0,2pi} is the function f(x) = tan(x) defined? Express your answer as a collection of intervals separated by commas.

My Answers: Cos 90degrees = 0;cos pi/2 = 0; sin 2pi = 0

I feel like I am missing something...Can someone please put this in baby steps for me! Thanks

skeeter · #5 $Old$ November 17th, 2008, 07:33 PM

$\tan{x} = \frac{\sin{x}}{\cos{x}}$

$\tan{x}$ is undefined anywhere $\cos{x} = 0$

on the interval $[0, 2\pi]$ , $\cos{x} = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$

next time, start a new question with a new post.

Atav20 · #6 $Old$ December 1st, 2008, 06:36 PM

Quote:

Originally Posted by skeeter $View Post$

your solutions are correct

for $y = A\cos[B(x+C)] + D$

$|A|$ is the amplitude

$\frac{2\pi}{|B|}$ is the period

$C$ is the phase shift ... (+) left , (-) right

$D$ is the vertical shift ... (+) up, (-) down

btw ... for future reference, the Greek letter $\pi$ is spelled "pi".

this is a piece of "pie".

Why is the phase shift -pi/15 and not pi/3 for the first problem

skeeter · #7 $Old$ December 1st, 2008, 06:47 PM

this is why ...

$f(x) = -\cos\left[5\left(x + \frac{\pi}{15}\right)\right]$