Trigonometric Equations

takuto · #1 $Old$ November 4th, 2009, 02:13 PM

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.

skeeter · #2 $Old$ November 4th, 2009, 02:44 PM

Quote:

Originally Posted by takuto $View Post$

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

takuto · #3 $Old$ November 4th, 2009, 03:24 PM

Quote:

Originally Posted by skeeter $View Post$

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

skeeter · #4 $Old$ November 4th, 2009, 03:50 PM

Quote:

Originally Posted by takuto $View Post$

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

satis · #5 $Old$ November 4th, 2009, 06:11 PM

Quote:

Originally Posted by takuto $View Post$

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.

pacman · #6 $Old$ November 5th, 2009, 09:30 PM

here is the graph of the two trigonometric equations, try to figure out the answer by inspection . . .