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Addition of sine waves 
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#19
Mar1114, 03:54 PM

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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. 


#20
Mar1114, 05:35 PM

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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. 


#21
Mar1214, 01:55 AM

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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}=C^{2}(sin^{2}(γ)+C^{2}cos^{2}(γ)=C^{2} You can expand and simplify the left hand side: A^{2}+B^{2}+2AB cos(β)=C^{2}. 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 


#22
Mar1214, 04:15 AM

P: 11

OK think I have finally understood. The answer I have come up with is:
Vc = 2.0007sin(314.2t59.99°) Thank you all for your help and if someone would confirm my answer for my peace of mind that would be great. 


#23
Mar1214, 06:38 AM

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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 


#24
Mar1214, 08:58 AM

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Thank you ehild



#25
Mar1214, 01:51 PM

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



#26
Mar1214, 02:05 PM

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



#27
Mar1214, 06:17 PM

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#28
Mar1314, 01:36 AM

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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 highschools. 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=j...62922401,d.bGQ ehild 


#29
Mar1314, 09:45 AM

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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. 


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