Combining Sine Waves: Adding 2 sin (314.2t) and 2 sin(314.2t - 120°)

[Maths dunce]

I need to add together the following sine waves and express the answer in the same form:

Va = 2 sin (314.2t) + Vb = 2 sin(314.2t - 120°)

Any help would be greatly appreciated.

 

[arildno]

Va = 2 sin (314.2t) + Vb = 2 sin(314.2t - 120°)

Not until you correct such sloppiness in writing.

 

[arildno]

Furthermore, do NOT make approx like 100*pi=314.2. I'm sure your book wrote 100*pi??

 

[Maths dunce]

My book stated the question exactly as I wrote it on here so I guess the author is sloppy and making approximations.

This is what really annoys people about forums lots of arsey know-it-alls who don't want to help just to show off how clever they are. Well done you

 

[arildno]

Well, if he DID write Va = 2 sin (314.2t) + Vb = 2 sin(314.2t - 120°), what does that tell you Va equals?

 

[Maths dunce]

I am sorry if I have offended you with my lack of knowledge or how the question was laid out for me. I just wanted a little bit of help with a maths problem and thought I might get some here.
I have two expressions Va = 2 sin (314.2t) and Vb = 2 sin (314.2t - 120°) and I need to know how to add them together and express the answer in the same form.
I do not know how to do this as I have never been taught so if you would like to help me that would be fantastic however if you would prefer to belittle me some more please feel free.

 

[arildno]

"I have two expressions Va = 2 sin (314.2t) and Vb = 2 sin (314.2t - 120°)"

THAT is how you should write it!

What you WROTE however, "Va = 2 sin (314.2t) + Vb = 2 sin(314.2t - 120°)" means something completely different, namely:

1. Va=2sin(314.2t)+Vb
and
2. 2sin(314.2t)+Vb=2sin(314.2t-120deg)

This means that, for example, combining 1 and 2, Va=2sin(314.2t-120deg)
and using this in 1., you may find Vb=2sin(314.2t-120deg)-2sin(314.2t)

That isn't quite what you meant, was it?

So, it is not about me "being offended", but that you should become aware how CAREFUL you must be not to make sloppinesses (and, believe me, it is sloppiness which comes NATURALLY to all of us, becoming careful needs lots of work!)

Before proceeding, do you see my point now?

 

[Maths dunce]

I do see your point and it was a valid one. I was frustrated after hours of fruitless searching on the internet for an answer to this question but as you say there is no excuse for sloppiness as it just causes confusion.

 

[DrewD]

So as it stands, your problem is finding how to rewrite

$2\sin(314.5t)+2\sin(314.5t-120)$

as a single term? If that is the case, there is an identity that allows you to combine such terms. It is near the bottom of the page here. If that is not what you are looking for, or you are still struggling to find the identity on that page, mention it.

PS this should probably be in "homework help"

 

[arildno]

To give you a clue:
If you had Sin(A)+Sin(B), could you make this into a simplified product expression?

 

[arildno]

I do see your point and it was a valid one. I was frustrated after hours of fruitless searching on the internet for an answer to this question but as you say there is no excuse for sloppiness as it just causes confusion.

Sloppiness occurs naturally, and there is no shame in that; the important thing is not to become of self-defensive if somebody tells you of it, but rather take a step back and think: "Hmm..perhaps I was sloppy? Let me just check one more time.."

In that manner, actual INSTANCES of sloppiness will decrease fast, because you let the ambition to always be most critical to yourself guide your work, acknowledging that in the beginning, others' criticisms might be beneficial spurs to develop the actually very difficult skill of being self-critical of work you have invested a lot of work in.

See if the advices given you of Sin(A)+Sin(B) leads you somewhere, hmm?

 

[Mark44]

I need to add together the following sine waves and express the answer in the same form:

Va = 2 sin (314.2t) + Vb = 2 sin(314.2t - 120°)

Any help would be greatly appreciated.

Using "+" to represent "and" was very confusing to the folks responding in this thread. What you wrote says that Va equals 2 sin (314.2t) + Vb, which in turn is equal to 2 sin(314.2t - 120°).

 

[Maths dunce]

Thank you for your responses. I have to go to work now but I will look into what you have suggested when I return. I appreciate what you have said arildno and will take it on board.

DrewD I am new to this forum and only realized after I posted that it was in the wrong section. I will be more careful in future.

 

[Maths dunce]

Just spent another two hours trying to get my head around this. I am sure I will appear to be pretty dumb to you clever people.
I have drawn the two sine waves on a graph and created a table of values, I have added these together and plotted the resultant wave. This has confused me even more because when I use any of the equations I have found online to calculate a given point it does not match. This says to me that I have miscalculated my points (which I don't think is the case as I have checked and double checked) or I still haven't found the right information.
If anyone can give me a bit more of a clue I would be most grateful.

 

[arildno]

Se Va=2*sin(A), Vb=2*sin(B), where A=314.2t, B=314.2t-120

Now, simplify the expression Va+Vb=2*(Sin(A)+Sin(B)) using a trigonometric identity.

 

[DrewD]

I have drawn the two sine waves on a graph and created a table of values, I have added these together and plotted the resultant wave.

This is a very nice bit of graphing practice, but don't do that. If you want, you can use a graphing calculator or online grapher like desmos.com, but I would not waste my time graphing by hand except for data.

This has confused me even more because when I use any of the equations I have found online to calculate a given point it does not match. This says to me that I have miscalculated my points (which I don't think is the case as I have checked and double checked) or I still haven't found the right information.

What information have you found? arildno simplified the problem for you. Which identity would you use?

 

[Maths dunce]

Thank you arildno for your help and your criticism. I have now been looking into this for over eight hours all in all and I am sure I am missing something very obvious but I just don't get it.
Unfortunately I work full time and have family responsibilities and I don't have any more time to dedicate to this. So I am officially throwing the towel.

 

[arildno]

Have you seen the math identity:
$$\sin(u)+\sin(v)=2sin(\frac{u+v}{2})\cos(\frac{u-v}{2})$$

use that one, with A=u and B=v.

Afterwards, insert your original expressions for A and B and simplify.

 

[Maths dunce]

I have seen this identity and if I insert my figures it does confirm that my addition of the points is correct. It is the simplifying / putting it back into the original form that I cannot understand. I will look into this further and get back to you.
Thank you for your patience.

 

[arildno]

Can you please post what you have done further?
Honestly, I do not see what problems you have reached into, and I am actually inclined to believe you have misunderstood the right answer as somehow not being what you were asked for.

 

[ehild]

So as it stands, your problem is finding how to rewrite

$2\sin(314.5t)+2\sin(314.5t-120)$

as a single term? If that is the case, there is an identity that allows you to combine such terms. It is near the bottom of the page ="http://www.sosmath.com/trig/Trig5/trig5/trig5.html"]here. If that is not what you are looking for, or you are still struggling to find the identity on that page, mention it.

PS this should probably be in "homework help"

You have to write the sum of two sine (cosine) functions of form V_a(t)=Asin(wt) and V_b(t)=Bsin(wt+β) as V_c(t)=C sin(wt+γ). For that, expand both Vb and Vc, using the identity

sin(a+b)=sin(a) cos(b) + cos(a) sin(b)

(or cos(a+b)=cos(a)cos(b)-sin(a)sin(b))

You get :

V_b= B [sin(wt)cos(β)+cos(wt)sin(β)] and V_c=C sin(wt+γ)=C[sin(wt)cos(γ)+cos(wt)sin(γ)].

With those, the equation V_a+V_b = V_c becomes

Asin(wt)+B sin(wt)cos(β)+Bcos(wt)sin(β)=Csin(wt)cos(γ)+Ccos(wt)sin(γ).

Collect all the sin(wt) terms and also the cos(wt) terms:

sin(wt)[A+Bcos(β)-Ccos(γ)] + cos(wt)[Bsin(β)-Csin(γ)]= 0

You have to find the unknown C and γ so that the equation holds at any time t.

For t=0, sin(wt)=0 cos(wt)=±1, so Bsin(β)-Csin(γ)=0.

If wt=pi/2 radian (t=pi/(2w) cos(wt)=0 and sin(wt)=±1, so A+Bcos(β)-Ccos(γ)=0.You have to solve the system of equation in bold.

Bsin(β)=Csin(γ)

A+Bcos(β)=Ccos(γ)

Square both equations and add them together.

(Bsin(β))²+(A+Bcos(β))²=C²(sin²(γ)+C²cos²(γ)=C²

You can expand and simplify the left hand side:

A²+B²+2AB cos(β)=C².

Knowing C, sin(γ)=B/C and cos(γ)=(A+Bcos(β))/C.

Apply to your problem: A=B=2, β=-120 degrees. (:Edited)

ehild

 

[Maths dunce]

OK think I have finally understood. The answer I have come up with is:

Vc = 2.0007sin(314.2t-59.99°)

Thank you all for your help and if someone would confirm my answer for my peace of mind that would be great.

 

[ehild]

It is correct but the rounding errors. Vc=2.000 sin(314.2t - 60.00°)

(I had an error in my previous post, beta was -120° instead of 120°.)ehild

 

[Maths dunce]

Thank you ehild

 

[arildno]

Going the path along with the half angle product formula would have given you precisely the same expression.

 

[Maths dunce]

Thank you arildno I understand that now. It took a while but I got there in the end.

 

[DrewD]

Going the path along with the half angle product formula would have given you precisely the same expression.

With a lot less work!

 

[ehild]

Yes, it is a lot less work if you remember that such formula exists.
The other method uses the basic addition laws of Trigonometry, taught and explained in detail in high schools.
You can also find the "ready" formulae for addition of two sinusoidal function of time, and it is also taught, why not use those? http://www.google.com/url?sa=t&rct=...=Vd_rop1AaDXfMkyVtL7PKA&bvm=bv.62922401,d.bGQ

ehild

 

[arildno]

Thank you arildno I understand that now. It took a while but I got there in the end.

However, I would like to say that ehild's scheme, although apparently clunky, is a VERY important technique in many other fields as well, i.e, understanding and using the concept of linearly independent functions in order to solve particular problems.

It is, in a manner of speaking, a generally more powerful tool than trig identity tables that you'll gradually learn to appreciate.