Question

I know how to graph a cosine equation in degrees, but how do I graph a sine equation in degrees?

Answer 1

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?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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