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anonymous
 3 years 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)
anonymous
 3 years 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)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1do you know how to expand sin(Bt+C)?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1using the angle sum formulae?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1@brazellnicole25 I can only help you if you respond to my questions.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes i have done that before

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay is it sin(a+B)=sinacosb+cos(a)sin(b)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yes  that is the correct formulae to use here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does it equal the equation that I wrote before? I am stuck here for some reason

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This 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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1no 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?

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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1great, 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)

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1can you now infer what the value of B must be?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1compare the term sin(Bt) and sin(t/2) and also cos(Bt) and cos(t/2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0b must be t is that right?

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1divide both sides by 't'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you write \(t/2\) as \((1/2)t\) you should be able to infer the value of \(B\)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1we have B times 't' is equal to (1/2) times 't' so what must B equal?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1ok, 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)

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

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

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

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

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1correct, 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:\[16+9=A^2\cos^2C)+A^2\sin^2(C)\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it does your just combining the two to make one equation.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1correct  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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1we 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes that does. you solved for c.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow i made that harder than it had to be for sure.

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I hope you understood all the steps

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm, i think C should be 18036.87.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I thank you so much for your time.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1yw :) @sirm3d  let me check what you are saying...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1both are correct @sirm3d since:\[\sin(x)=\sin(180x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y_2=5 \sin (\frac{1}{2}t +143.17)\] because \(4=5 \cos C\) abd there's no way to reconcile the sign if C is in quadrant I

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I believe you are right @sirm3d since: 4 = Acos(C)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1which implies cos(C) must be negative thank you for correcting me. :)

asnaseer
 3 years ago
Best 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dividing the two equations should yield an equation in tangent.\[\frac{3}{4}=\frac{\sin C}{\cos C}=\tan C\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I hope that is also clear to you @brazellnicole25

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay yeah so you combined the two right? and then solved for C. is that correct

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1so 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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok I got you. So can I can I ask you something about this if you have time?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1if its related to this question then sure

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y1 and y2 would be the same equations right? I mean if I was to graph them?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay thank you so much. I really appriciate it. and your time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could i ask you another question about this?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1BTW: 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1not at all  I am glad you find the subject interesting :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and apply things I've already learned in class. That's why i'm asking so much.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I do love math and I like to do it more when I understand it more.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank 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.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1again, 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. :)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0^^ thanks, @asnaseer i teach math so i actually get to see the ins and outs of most mathematics.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1what age group do you teach @sirm3d ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0college math. up to engineering mathematics, @asnaseer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I like sirm3d how you helped us to.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1thats 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.... :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i love math, and i find joy seeing students learn and appreciate math.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay asnaseer I hope you have a good night. thanks alot.
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