# Trigonometric Equations

• Nov 4th 2009, 01:13 PM
takuto
Trigonometric Equations
Hi. I need help to find the:
-Amplitude
-Period
-Vertical Shift
-Horizontal Shift

1) y= 2sin(4x+pi)+3
2) y= -3cos((x/2)+2pi)-5

Any help would be greatly appreciated.
• Nov 4th 2009, 01:44 PM
skeeter
Quote:

Originally Posted by takuto
Hi. I need help to find the:
-Amplitude
-Period
-Vertical Shift
-Horizontal Shift

1) y= 2sin(4x+pi)+3
2) y= -3cos((x/2)+2pi)-5

Any help would be greatly appreciated.

$y = A\cos[B(x + C)] + D$ ... also sine

$|A|$ = amplitude

period = $\frac{2\pi}{|B|}$

$C$ = horizontal shift ... (+) left , (-) right

$D$ = vertical shift ... (+) up , (-) down
• Nov 4th 2009, 02:24 PM
takuto
Quote:

Originally Posted by skeeter
$y = A\cos[B(x + C)] + D$ ... also sine

$|A|$ = amplitude

period = $\frac{2\pi}{|B|}$

$C$ = horizontal shift ... (+) left , (-) right

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

I was wondering if I did it right:
2sin[4x+pi]+3
A=2
B=pi/2
C=pi/4
D=3

-3cos((x/2)+2pi))-5
A=-3
B=4pi
C=0
D=-5
• Nov 4th 2009, 02:50 PM
skeeter
Quote:

Originally Posted by takuto
I was wondering if I did it right:
2sin[4x+pi]+3
A=2
B=pi/2
C=pi/4
D=3

-3cos((x/2)+2pi))-5
A=-3
B=4pi
C=0
D=-5

no
• Nov 4th 2009, 05:11 PM
satis
Quote:

Originally Posted by takuto
I was wondering if I did it right:
2sin[4x+pi]+3
A=2
B=pi/2
C=pi/4
D=3

-3cos((x/2)+2pi))-5
A=-3
B=4pi
C=0
D=-5

a few things to remember

Amplitude is never negative (it's an absolute value)
the period of a sin and cos is 2pi / B. Same for sec and csc,
vertical shift is the number that's outside (you got that right in these cases)
horizontal shift is -C/B if I'm not mistaken.
• Nov 5th 2009, 08:30 PM
pacman
here is the graph of the two trigonometric equations, try to figure out the answer by inspection . . .