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

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metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1I 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?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0upper and lower bounds of sin x are 1 and 1 and sin 0 = 0 so the curve goes through the origin

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1I know that, i'm just confused on the graphing of it. For example: Y=20sin1/2(x120 degree) 10

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1If it was a cosine equation, I could just graph the point on (120 deg, 10) and that would be the upper bound point.

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1But 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y = sin x passes through the origin. Use the phase shift value to shift the function horizontally. If that's what you're asking

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1So basically were on x would the point be located at? Since the phase shift is 120, should I do 12090 or 120+90

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1I can attach the book answer, the graph approximately starts at 180 degrees horizontally. Even thought the phase shift is 120

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not sure where the 90 is coming from. Shift it 120° to the right This is y = sin xdw:1442167291189:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this would be sin (x  120°) dw:1442167349251:dw

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1yes I figured that too, but please look the the image, I will attatch it now

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1Problem: 3. Again the equation is: 20sin1/2(x120deg) 10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. 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)

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1The point that the graph passes through is 180 not 120. Or is it 120?

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1nevermind, I see it

metalslayer
 one year ago
Best ResponseYou've already chosen the best response.1Oh that makes sense now

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i recall this as \[A\sin(\psi(t\phi))+B\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, so the point that is at the origin in y = sin x is translated to \((\phi,B)\)
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