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Can anyone help me with a trig problem. I need to turn y1=4sin(t/2)+3cos(t/2) into the form y2=Asin(Bt+C)
 one year ago
 one year ago
Can anyone help me with a trig problem. I need to turn y1=4sin(t/2)+3cos(t/2) into the form y2=Asin(Bt+C)
 one year ago
 one year ago

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brazellnicole25Best ResponseYou've already chosen the best response.0
I would like to know how to obtain the answer step by step. I don't want to just know the answer. I would like to understand how it was obtained.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
do you know how to expand sin(Bt+C)?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
using the angle sum formulae?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
@brazellnicole25 I can only help you if you respond to my questions.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
I will repond. i know i'm suppost to use the transformation identity to make the first equation the other one. and c is suppost to be the smallest positive ange.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
first expand sin(Bt+C) using the angle sum formulae  do you know how to do that?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
i.e do you know how to express sin(x+y) in terms of sin(x) and cos(x)?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
and sin(y) and cos(y)
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes i have done that before
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
ok, good, so please first write sin(Bt+C) in its expanded form
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
okay is it sin(a+B)=sinacosb+cos(a)sin(b)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes  that is the correct formulae to use here
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
does it equal the equation that I wrote before? I am stuck here for some reason
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
ok, let me take you through this one stepbystep and then hopefully you will be able to do others that are similar to this one
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
This is where I'm having my block for this problem for some reason. I've used the above before but I dont' know why i'm having so much trouble. I'm sorry I'm really trying
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
no problem  lets work through it together. we start by first expanding sin(Bt+C) to get: sin(Bt+C) = sin(Bt)cos(C) + cos(Bt)sin(C) agreed?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
good, next we multiply this by A to get: Asin(Bt+C) = Asin(Bt)cos(C) + Acos(Bt)sin(C) agreed?
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes because we have to do the same to all of them (distribute) okay i'm following
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
great, now lets rearrange this to get it into a similar form to the original equation. we therefore get: Asin(Bt+C) = Acos(C) * sin(Bt) + Asin(C) * cos(Bt)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
compare this to the original equation which is: 4*sin(t/2) + 3*cos(t/2)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
can you now infer what the value of B must be?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
compare the term sin(Bt) and sin(t/2) and also cos(Bt) and cos(t/2)
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
b must be t is that right?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
not quite  if you compare these terms then we see that: Bt = t/2 therefore B = ?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
divide both sides by 't'
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
if you write \(t/2\) as \((1/2)t\) you should be able to infer the value of \(B\)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
we have B times 't' is equal to (1/2) times 't' so what must B equal?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
ok, so now we can use this to get: 4*sin(t/2) + 3*cos(t/2) = Acos(C) * sin(t/2) + Asin(C) * cos(t/2)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
now compare the terms that are multiplying the sin(t/2) terms and the cos(t/2) terms on both sides of the equals sign
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
can you see that this means that: 4 = Acos(C) 3 = Asin(C)
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes I can see that. The right side of the equal sign it just streched out. yeah that makes sense
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
good, next are you aware of the identity:\[\sin^2(x)+\cos^2(x)=1\]
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes and i know that can me manipulated in a couple of ways. that is one of the pythagorean identies
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
correct, so if we use this identity, we can square both sides of: 4 = Acos(C) 3 = Asin(C) to get:\[16 = A^2\cos^2(C)\]\[9=A^2\sin^2(C)\]and then add them together to get:\[16+9=A^2\cos^2C)+A^2\sin^2(C)\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
therefore:\[25=A^2(\cos^2(C)+\sin^2(C))=A^2(1)=A^2\]does that make sense?
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes it does your just combining the two to make one equation.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
correct  now we take square roots of both sides to get:\[A=\sqrt{25}=5\]so now we have found A=5 and B=1/2 which just leaves C to find
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yeah that's right
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
we can find C using either one of the two earlier equations we got to: 4 = Acos(C) 3 = Asin(C) lets use the second one to get: 3 = Asin(C) therefore: sin(C) = 3/A = 3/5 therefore: C = arcsin(3/5) = 36.87 degrees (approximately)
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes that does. you solved for c.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
wow i made that harder than it had to be for sure.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
so we can finally write: y2 = 5sin(t/2 + 36.87) or, if you prefer: y2 = 5sin(t/2 + arcsin(3/5))
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I hope you understood all the steps
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes I did. Thank you so much. I really appriciate it. I was looking back at some of these problems that I missed on a previous test and it was bothering me so I wanted to figure them out. Just so I would know. I don't want to be behind in my calculus class next semester.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
hmmm, i think C should be 18036.87.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
I thank you so much for your time.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yw :) @sirm3d  let me check what you are saying...
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
both are correct @sirm3d since:\[\sin(x)=\sin(180x)\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[y_2=5 \sin (\frac{1}{2}t +143.17)\] because \(4=5 \cos C\) abd there's no way to reconcile the sign if C is in quadrant I
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I believe you are right @sirm3d since: 4 = Acos(C)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
which implies cos(C) must be negative thank you for correcting me. :)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
@brazellnicole25  to avoid this type of mistake, what we should have done is gone from these two equations: 4 = Acos(C) 3 = Asin(C) to this one by diving the two: tan(C) = 3/4 that guarantees that we get the correct quadrant.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
dividing the two equations should yield an equation in tangent.\[\frac{3}{4}=\frac{\sin C}{\cos C}=\tan C\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I hope that is also clear to you @brazellnicole25
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
okay yeah so you combined the two right? and then solved for C. is that correct
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes because the angle C has to satisfy BOTH equations: 4 = Acos(C) 3 = Asin(C)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
so we just divided the second one by the first one to get:\[\frac{3}{4}=\frac{A\sin(C)}{A\cos(C)}\]which lead to: tan(C) = 3/4
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
so the correct solution is: y2 = 5sin(t/2  arctan(3/4))
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
sorry: y2 = 5sin(t/2 + arctan(3/4))
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
ok I got you. So can I can I ask you something about this if you have time?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
if its related to this question then sure
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
y1 and y2 would be the same equations right? I mean if I was to graph them?
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
yes it is about this one. sorry i ment to put that first.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
okay thank you so much. I really appriciate it. and your time.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
I haven't ever used this site is there anything I can do to help you get a beter rating or anything? I don't know alot about this site.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I am glad I was able to help  I do this because I enjoy teaching those who are willing to learn, not for any rating. But, in general, the usual workflow is: 1. You post a question. 2. One or more people come to your question to help you out. 3. You select the one that best answered your question by clicking on the blue Best Response button next to any one of their replies. 4. You close the question by clicking on the Close button then the whole process starts again.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
could i ask you another question about this?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
BTW: You can use wolframalpha to see the graphs as follows: y1: http://www.wolframalpha.com/input/?i=y%3D4sin%28t%2F2%29%2B3cos%28t%2F2%29 y2: http://www.wolframalpha.com/input/?i=y+%3D+5sin%28t%2F2+%2B+2.4981%29
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
would the amplitude of this be 5 , the period be 2pie (symbol i mean) and the phanse shift be .... sorry i'm thinking about that for a min.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
okay thanks for showing me the graphs. I got the same graphs on my calculator at on these you gave me. Thank you so much for that.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
you can learn more about phase shift and amplitudes here: http://www.purplemath.com/modules/grphtrig.htm
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
in general:\[Y=A\sin(wt+\phi)\]means: A=amplitude \(w=2*\pi/P\) P= period \(\phi\)=phase shift
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
okay sorry to be so full of questions. I'm just rying to fully understand it. I hope i'm not being to much trouble. I really do appriciate it though.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
not at all  I am glad you find the subject interesting :)
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
and apply things I've already learned in class. That's why i'm asking so much.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
I do love math and I like to do it more when I understand it more.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
it is the same with any subject  if you really enjoy a subject then you will naturally get better at it.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
a thirst for knowledge is always a good sign  I am sure you will do really well and achieve all that you want to achieve. :)
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
Thank you so much. I really appriciate it. Your very smart though. do you teach math somewhere? Not that I mean to be to nosey. Your good at it though.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
again, you are more than welcome. And no I am not an actual maths teacher, I work as a fulltime software engineer but always enjoyed maths since high school, so I try and keep up with the subject by /trying/ to teach others. :)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
I have also learnt a LOT of things from other on this site  @sirm3d being one of them. :)
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
^^ thanks, @asnaseer i teach math so i actually get to see the ins and outs of most mathematics.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
what age group do you teach @sirm3d ?
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
well you are good I mean great at it. I like how you take the time to help others and make sure the person your helping will interact with you. if they really want to learn then they would have no problem answering you.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
college math. up to engineering mathematics, @asnaseer
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
I like sirm3d how you helped us to.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
thats good to know @sirm3d  I'll keep you in mind if I run into any problems that I get stuck in. ok folks  I have to go now as it is 1:40am where I am and I need to catch some Zzzz's.... :)
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
i love math, and i find joy seeing students learn and appreciate math.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
you were very polite in giving your input. I have gone to get help before and some people are just so rude. but your so nice about helping.
 one year ago

brazellnicole25Best ResponseYou've already chosen the best response.0
okay asnaseer I hope you have a good night. thanks alot.
 one year ago
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