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do you know how to expand sin(Bt+C)?

using the angle sum formulae?

@brazellnicole25 I can only help you if you respond to my questions.

first expand sin(Bt+C) using the angle sum formulae - do you know how to do that?

i.e do you know how to express sin(x+y) in terms of sin(x) and cos(x)?

and sin(y) and cos(y)

yes i have done that before

ok, good, so please first write sin(Bt+C) in its expanded form

okay is it sin(a+-B)=sinacosb+-cos(a)sin(b)

yes - that is the correct formulae to use here

okay.

so sin(Bt+C)=?

does it equal the equation that I wrote before? I am stuck here for some reason

yes

good, next we multiply this by A to get:
Asin(Bt+C) = Asin(Bt)cos(C) + Acos(Bt)sin(C)
agreed?

yes because we have to do the same to all of them (distribute) okay i'm following

compare this to the original equation which is:
-4*sin(t/2) + 3*cos(t/2)

can you now infer what the value of B must be?

compare the term sin(Bt) and sin(t/2)
and also cos(Bt) and cos(t/2)

b must be t is that right?

t/2

not quite - if you compare these terms then we see that:
Bt = t/2
therefore B = ?

divide both sides by 't'

if you write \(t/2\) as \((1/2)t\) you should be able to infer the value of \(B\)

we have B times 't' is equal to (1/2) times 't'
so what must B equal?

1/2

perfect!

can you see that this means that:
-4 = Acos(C)
3 = Asin(C)

yes I can see that. The right side of the equal sign it just streched out. yeah that makes sense

good, next are you aware of the identity:\[\sin^2(x)+\cos^2(x)=1\]

yes and i know that can me manipulated in a couple of ways. that is one of the pythagorean identies

therefore:\[25=A^2(\cos^2(C)+\sin^2(C))=A^2(1)=A^2\]does that make sense?

yes it does your just combining the two to make one equation.

yeah that's right

make sense?

yes that does. you solved for c.

wow i made that harder than it had to be for sure.

so we can finally write:
y2 = 5sin(t/2 + 36.87)
or, if you prefer:
y2 = 5sin(t/2 + arcsin(3/5))

I hope you understood all the steps

hmmm, i think C should be 180-36.87.

I thank you so much for your time.

actually no

which implies cos(C) must be negative
thank you for correcting me. :)

exactly

I hope that is also clear to you @brazellnicole25

okay yeah so you combined the two right? and then solved for C. is that correct

yes because the angle C has to satisfy BOTH equations:
-4 = Acos(C)
3 = Asin(C)

so the correct solution is:
y2 = 5sin(t/2 - arctan(3/4))

sorry:
y2 = 5sin(t/2 + arctan(-3/4))

ok I got you. So can I can I ask you something about this if you have time?

if its related to this question then sure

y1 and y2 would be the same equations right? I mean if I was to graph them?

yes

yes it is about this one. sorry i ment to put that first.

okay thank you so much. I really appriciate it. and your time.

could i ask you another question about this?

in general:\[Y=A\sin(wt+\phi)\]means:
A=amplitude
\(w=2*\pi/P\)
P= period
\(\phi\)=phase shift

not at all - I am glad you find the subject interesting :)

and apply things I've already learned in class. That's why i'm asking so much.

I do love math and I like to do it more when I understand it more.

*others

I like sirm3d how you helped us to.

i love math, and i find joy seeing students learn and appreciate math.

okay asnaseer I hope you have a good night. thanks alot.

night night... :)