## metalslayer one year ago I know how to graph a cosine equation in degrees, but how do I graph a sine equation in degrees?

1. metalslayer

I know how to graph a cosine equation. Basically the amplitude (minus/plus the vertical shift) is the upper bound, and I know how to get the lower bound. For sine equations I can find the upper lower bounds and the sinuoidal axis. BUT, HOW DO I FIND WERE TO EXACTLY START THE EQUATION ACCORDING TO THE PERIOD?

2. welshfella

upper and lower bounds of sin x are -1 and 1 and sin 0 = 0 so the curve goes through the origin

3. metalslayer

I know that, i'm just confused on the graphing of it. For example: Y=20sin1/2(x-120 degree) -10

4. metalslayer

If it was a cosine equation, I could just graph the point on (120 deg, 10) and that would be the upper bound point.

5. metalslayer

But for sine, were exactly should I start at the x axis? Basically I now the Y axis and how high/low it would be, but were do I start is my question? Thanks.

6. anonymous

y = sin x passes through the origin. Use the phase shift value to shift the function horizontally. If that's what you're asking

7. metalslayer

So basically were on x would the point be located at? Since the phase shift is 120, should I do 120-90 or 120+90

8. metalslayer

I can attach the book answer, the graph approximately starts at 180 degrees horizontally. Even thought the phase shift is 120

9. anonymous

Not sure where the 90 is coming from. Shift it 120° to the right This is y = sin x|dw:1442167291189:dw|

10. anonymous

so this would be sin (x - 120°) |dw:1442167349251:dw|

11. metalslayer

yes I figured that too, but please look the the image, I will attatch it now

12. anonymous

ok

13. metalslayer

14. metalslayer

Problem: 3. Again the equation is: 20sin1/2(x-120deg) -10

15. anonymous

ok. I see. The midline of the sine function passes through the origin. that's what's at 120°. so your first point should be (120°, -10)

16. metalslayer

The point that the graph passes through is 180 not 120. Or is it 120?

17. metalslayer

nevermind, I see it

18. metalslayer

Oh that makes sense now

19. amistre64

i recall this as $A\sin(\psi(t-\phi))+B$

20. anonymous

yes, so the point that is at the origin in y = sin x is translated to $$(\phi,B)$$