More trig trauma!

**MathBlaster47** · February 4th 2010, 06:22 PM

These questions get nothing but a blank stare from me....kinda irritating really since I sort of like trigonometry.

Describe each of the following graphs of y. Give (a) its amplitude (if applicable),(b) its maximum and minimum (if applicable) and (c) its period.
If it has no amplitude, maximum or minimum, write none in the appropriate space.

1
$y=3\sin\frac{1}{3}x$
(a) 3(I think)

(b)no idea

(c) $\frac{2\pi}{\frac{1}{3}}$ (I hope)

2
$y=\tan 2A$

(a) No clue

(b) Maximum:none and Minimum: none

(c) I wish I knew...

Thank you muchly once again!!

**pickslides** · February 4th 2010, 06:26 PM

Here's a kick start for the period.

Period for $\sin(bx)$ is $\frac{2\pi}{b}$

Period for $\tan(bx)$ is $\frac{\pi}{b}$

**MathBlaster47** · February 4th 2010, 06:33 PM

Originally Posted by pickslides

Here's a kick start for the period.

Period for $\sin(bx)$ is $\frac{2\pi}{b}$

Period for $\tan(bx)$ is $\frac{\pi}{b}$

Ok....so the period for $y=3\sin\frac{1}{3}x$ is $\frac{2\pi}{\frac{1}{3}}$

and the period for [/COLOR][/COLOR] $y=\tan 2A$ is $\frac{\pi}{2}$ ?

**mattio** · February 4th 2010, 07:56 PM

Originally Posted by MathBlaster47

These questions get nothing but a blank stare from me....kinda irritating really since I sort of like trigonometry.

Describe each of the following graphs of y. Give (a) its amplitude (if applicable),(b) its maximum and minimum (if applicable) and (c) its period.
If it has no amplitude, maximum or minimum, write none in the appropriate space.

1
$y=3\sin\frac{1}{3}x$
(a) 3(I think)

(b)no idea

(c) $\frac{2\pi}{\frac{1}{3}}$ (I hope)

2
$y=\tan 2A$

(a) No clue

(b) Maximum:none and Minimum: none

(c) I wish I knew...

Thank you muchly once again!!

#1

A. amplitude is 3
B. since graph has no vertical shift and amp is 3, minimum is -3 & max is 3
c. period is 2pi/b, our b is 1/3 so period is 6pi

#2

A. amplitude is understood to be 1
B. thus, the minimum is -1 and the maximum is 1
C. period is pi/b, so pi/2

**pickslides** · February 7th 2010, 11:51 AM

Originally Posted by MathBlaster47

Ok....so the period for $y=3\sin\frac{1}{3}x$ is $\frac{2\pi}{\frac{1}{3}}$

and the period for [/color][/color] $y=\tan 2A$ is $\frac{\pi}{2}$ ?

Yep and yep.

**mathemagister** · February 15th 2010, 02:37 PM

Hey MathBlaster47! Please keep in mind though, that the amplitude is the absolute value of the coefficient of the trigonometric term.

So if your first question was $y=-3\sin\frac{1}{3}x$ , the amplitude would still be 3.

Just keep that in mind, that is a clever little trap that most tests use to catch students off guard. (I've seen this trap in every big trig test I took.)

Hope that helps you in your future endeavors

**skeeter** · February 15th 2010, 04:46 PM

Originally Posted by mattio

#1

A. amplitude is 3
B. since graph has no vertical shift and amp is 3, minimum is -3 & max is 3
c. period is 2pi/b, our b is 1/3 so period is 6pi

#2

A. amplitude is understood to be 1 no
B. thus, the minimum is -1 and the maximum is 1 no
C. period is pi/b, so pi/2

the tangent function has no amplitude. it also has no maximum and minimum value since its range is $(-\infty , \infty)$

**mathemagister** · February 16th 2010, 12:28 AM

That is also a good catch, Skeeter. So, to sum up what Skeeter and I have said about amplitudes.

The amplitude of $-3sinx$ is 3.

The amplitude of $3cosx$ is 3.

The amplitude of $-3tanx$ does not exist.

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