metalslayer
  • metalslayer
I know how to graph a cosine equation in degrees, but how do I graph a sine equation in degrees?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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metalslayer
  • 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?
welshfella
  • welshfella
upper and lower bounds of sin x are -1 and 1 and sin 0 = 0 so the curve goes through the origin
metalslayer
  • metalslayer
I know that, i'm just confused on the graphing of it. For example: Y=20sin1/2(x-120 degree) -10

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metalslayer
  • 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.
metalslayer
  • 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.
anonymous
  • 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
metalslayer
  • 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
metalslayer
  • metalslayer
I can attach the book answer, the graph approximately starts at 180 degrees horizontally. Even thought the phase shift is 120
anonymous
  • anonymous
Not sure where the 90 is coming from. Shift it 120° to the right This is y = sin x|dw:1442167291189:dw|
anonymous
  • anonymous
so this would be sin (x - 120°) |dw:1442167349251:dw|
metalslayer
  • metalslayer
yes I figured that too, but please look the the image, I will attatch it now
anonymous
  • anonymous
ok
metalslayer
  • metalslayer
1 Attachment
metalslayer
  • metalslayer
Problem: 3. Again the equation is: 20sin1/2(x-120deg) -10
anonymous
  • 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)
metalslayer
  • metalslayer
The point that the graph passes through is 180 not 120. Or is it 120?
metalslayer
  • metalslayer
nevermind, I see it
metalslayer
  • metalslayer
Oh that makes sense now
amistre64
  • amistre64
i recall this as \[A\sin(\psi(t-\phi))+B\]
anonymous
  • anonymous
yes, so the point that is at the origin in y = sin x is translated to \((\phi,B)\)

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