anonymous
  • anonymous
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)
Trigonometry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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.
asnaseer
  • asnaseer
do you know how to expand sin(Bt+C)?
asnaseer
  • asnaseer
using the angle sum formulae?

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asnaseer
  • asnaseer
@brazellnicole25 I can only help you if you respond to my questions.
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
first expand sin(Bt+C) using the angle sum formulae - do you know how to do that?
asnaseer
  • asnaseer
i.e do you know how to express sin(x+y) in terms of sin(x) and cos(x)?
asnaseer
  • asnaseer
and sin(y) and cos(y)
anonymous
  • anonymous
yes i have done that before
asnaseer
  • asnaseer
ok, good, so please first write sin(Bt+C) in its expanded form
anonymous
  • anonymous
okay is it sin(a+-B)=sinacosb+-cos(a)sin(b)
asnaseer
  • asnaseer
yes - that is the correct formulae to use here
anonymous
  • anonymous
okay.
asnaseer
  • asnaseer
so sin(Bt+C)=?
anonymous
  • anonymous
does it equal the equation that I wrote before? I am stuck here for some reason
asnaseer
  • asnaseer
ok, let me take you through this one step-by-step and then hopefully you will be able to do others that are similar to this one
anonymous
  • anonymous
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
asnaseer
  • asnaseer
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?
anonymous
  • anonymous
yes
asnaseer
  • asnaseer
good, next we multiply this by A to get: Asin(Bt+C) = Asin(Bt)cos(C) + Acos(Bt)sin(C) agreed?
anonymous
  • anonymous
yes because we have to do the same to all of them (distribute) okay i'm following
asnaseer
  • asnaseer
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)
asnaseer
  • asnaseer
compare this to the original equation which is: -4*sin(t/2) + 3*cos(t/2)
asnaseer
  • asnaseer
can you now infer what the value of B must be?
asnaseer
  • asnaseer
compare the term sin(Bt) and sin(t/2) and also cos(Bt) and cos(t/2)
anonymous
  • anonymous
b must be t is that right?
anonymous
  • anonymous
t/2
asnaseer
  • asnaseer
not quite - if you compare these terms then we see that: Bt = t/2 therefore B = ?
asnaseer
  • asnaseer
divide both sides by 't'
sirm3d
  • sirm3d
if you write \(t/2\) as \((1/2)t\) you should be able to infer the value of \(B\)
anonymous
  • anonymous
2
asnaseer
  • asnaseer
we have B times 't' is equal to (1/2) times 't' so what must B equal?
anonymous
  • anonymous
1/2
asnaseer
  • asnaseer
perfect!
asnaseer
  • asnaseer
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)
asnaseer
  • asnaseer
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
asnaseer
  • asnaseer
can you see that this means that: -4 = Acos(C) 3 = Asin(C)
anonymous
  • anonymous
yes I can see that. The right side of the equal sign it just streched out. yeah that makes sense
asnaseer
  • asnaseer
good, next are you aware of the identity:\[\sin^2(x)+\cos^2(x)=1\]
anonymous
  • anonymous
yes and i know that can me manipulated in a couple of ways. that is one of the pythagorean identies
asnaseer
  • asnaseer
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:\[16+9=A^2\cos^2C)+A^2\sin^2(C)\]
asnaseer
  • asnaseer
therefore:\[25=A^2(\cos^2(C)+\sin^2(C))=A^2(1)=A^2\]does that make sense?
anonymous
  • anonymous
yes it does your just combining the two to make one equation.
asnaseer
  • asnaseer
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
anonymous
  • anonymous
yeah that's right
asnaseer
  • asnaseer
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)
asnaseer
  • asnaseer
make sense?
anonymous
  • anonymous
yes that does. you solved for c.
anonymous
  • anonymous
wow i made that harder than it had to be for sure.
asnaseer
  • asnaseer
so we can finally write: y2 = 5sin(t/2 + 36.87) or, if you prefer: y2 = 5sin(t/2 + arcsin(3/5))
asnaseer
  • asnaseer
I hope you understood all the steps
anonymous
  • anonymous
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.
sirm3d
  • sirm3d
hmmm, i think C should be 180-36.87.
anonymous
  • anonymous
I thank you so much for your time.
asnaseer
  • asnaseer
yw :) @sirm3d - let me check what you are saying...
asnaseer
  • asnaseer
both are correct @sirm3d since:\[\sin(x)=\sin(180-x)\]
asnaseer
  • asnaseer
actually no
sirm3d
  • sirm3d
\[y_2=5 \sin (\frac{1}{2}t +143.17)\] because \(-4=5 \cos C\) abd there's no way to reconcile the sign if C is in quadrant I
asnaseer
  • asnaseer
I believe you are right @sirm3d since: -4 = Acos(C)
asnaseer
  • asnaseer
which implies cos(C) must be negative thank you for correcting me. :)
asnaseer
  • asnaseer
@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.
sirm3d
  • sirm3d
dividing the two equations should yield an equation in tangent.\[\frac{3}{-4}=\frac{\sin C}{\cos C}=\tan C\]
asnaseer
  • asnaseer
exactly
asnaseer
  • asnaseer
I hope that is also clear to you @brazellnicole25
anonymous
  • anonymous
okay yeah so you combined the two right? and then solved for C. is that correct
asnaseer
  • asnaseer
yes because the angle C has to satisfy BOTH equations: -4 = Acos(C) 3 = Asin(C)
asnaseer
  • asnaseer
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
asnaseer
  • asnaseer
so the correct solution is: y2 = 5sin(t/2 - arctan(3/4))
asnaseer
  • asnaseer
sorry: y2 = 5sin(t/2 + arctan(-3/4))
anonymous
  • anonymous
ok I got you. So can I can I ask you something about this if you have time?
asnaseer
  • asnaseer
if its related to this question then sure
anonymous
  • anonymous
y1 and y2 would be the same equations right? I mean if I was to graph them?
asnaseer
  • asnaseer
yes
anonymous
  • anonymous
yes it is about this one. sorry i ment to put that first.
anonymous
  • anonymous
okay thank you so much. I really appriciate it. and your time.
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
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.
anonymous
  • anonymous
could i ask you another question about this?
asnaseer
  • asnaseer
BTW: You can use wolframalpha to see the graphs as follows: y1: http://www.wolframalpha.com/input/?i=y%3D-4sin%28t%2F2%29%2B3cos%28t%2F2%29 y2: http://www.wolframalpha.com/input/?i=y+%3D+5sin%28t%2F2+%2B+2.4981%29
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
you can learn more about phase shift and amplitudes here: http://www.purplemath.com/modules/grphtrig.htm
asnaseer
  • asnaseer
in general:\[Y=A\sin(wt+\phi)\]means: A=amplitude \(w=2*\pi/P\) P= period \(\phi\)=phase shift
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
not at all - I am glad you find the subject interesting :)
anonymous
  • anonymous
and apply things I've already learned in class. That's why i'm asking so much.
anonymous
  • anonymous
I do love math and I like to do it more when I understand it more.
asnaseer
  • asnaseer
it is the same with any subject - if you really enjoy a subject then you will naturally get better at it.
asnaseer
  • asnaseer
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. :)
anonymous
  • anonymous
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.
asnaseer
  • asnaseer
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. :)
asnaseer
  • asnaseer
I have also learnt a LOT of things from other on this site - @sirm3d being one of them. :)
asnaseer
  • asnaseer
*others
sirm3d
  • sirm3d
^^ thanks, @asnaseer i teach math so i actually get to see the ins and outs of most mathematics.
asnaseer
  • asnaseer
what age group do you teach @sirm3d ?
anonymous
  • anonymous
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.
sirm3d
  • sirm3d
college math. up to engineering mathematics, @asnaseer
anonymous
  • anonymous
I like sirm3d how you helped us to.
asnaseer
  • asnaseer
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.... :)
sirm3d
  • sirm3d
i love math, and i find joy seeing students learn and appreciate math.
anonymous
  • anonymous
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.
anonymous
  • anonymous
okay asnaseer I hope you have a good night. thanks alot.
asnaseer
  • asnaseer
night night... :)

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