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Mathematics 14 Online
rebeccaxhawaii (rebeccaxhawaii):

yay trig huehuehuehue

rebeccaxhawaii (rebeccaxhawaii):

wait. lemme get the question.

rebeccaxhawaii (rebeccaxhawaii):

is c always negative?

rebeccaxhawaii (rebeccaxhawaii):

phase shift c/b

satellite73 (satellite73):

oh dear c is probably a variable, so it could be positive or negative

rebeccaxhawaii (rebeccaxhawaii):

no but its|dw:1464142539442:dw| to find it right?

satellite73 (satellite73):

to find what?

rebeccaxhawaii (rebeccaxhawaii):

phase shift

satellite73 (satellite73):

phase shift of what?

rebeccaxhawaii (rebeccaxhawaii):

this is why screenshots are my best friend

satellite73 (satellite73):

how about a nice screen shot or light box or whatever those things are

rebeccaxhawaii (rebeccaxhawaii):

yee you already know

rebeccaxhawaii (rebeccaxhawaii):

http://prntscr.com/b83l4b looki

satellite73 (satellite73):

ok yikes this is my guess the phase shift for \[\sin(bx-c)\] is \[\frac{c}{b}\]

satellite73 (satellite73):

so for example the phase shift for \[\sin(2x-3)\] would be \(\frac{3}{2}\)

satellite73 (satellite73):

and the phase shift for \[\sin(4x+1)\] would be \[-\frac{1}{4}\]

rebeccaxhawaii (rebeccaxhawaii):

so its -c/b

satellite73 (satellite73):

look don't confuse yourself

satellite73 (satellite73):

they say it is \(\frac{c}{b}\) that is for \[\sin(bx-c)\]

satellite73 (satellite73):

for some unknown reason they put the minus sign here \[\huge \sin(bx\color{red}-c)\]

rebeccaxhawaii (rebeccaxhawaii):

don't confuse myself? that a little much to ask, you know thats my thing.

satellite73 (satellite73):

and said the phase shift was \(\frac{c}{b}\)

rebeccaxhawaii (rebeccaxhawaii):

thats*

rebeccaxhawaii (rebeccaxhawaii):

ohh okay so its just the opposite

satellite73 (satellite73):

but if they had writte \[\huge \sin(bx\color{blue}+c)\] then it would be \[-\frac{c}{b}\]

rebeccaxhawaii (rebeccaxhawaii):

OUI. Ill write that down. thanks chu i already know "yw"

satellite73 (satellite73):

here is the way to find it without memorizing anything

satellite73 (satellite73):

spose you got \[\cos(3x-7)\] and you want the phase shift solve \[3x-7=0\\ 3x=7\\ x=\frac{7}{3}\] and that is your phase shifty shift

satellite73 (satellite73):

yw

rebeccaxhawaii (rebeccaxhawaii):

called it huehuehue but frfr thank you

satellite73 (satellite73):

is that how they chortle in hawaii? huehuehue?

rebeccaxhawaii (rebeccaxhawaii):

rae jay watch this. you'll know what i mean. check pm.

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