# Addition of sine waves

by Maths dunce
Tags: addition, sine, waves
 P: 11 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.
 Sci Advisor HW Helper PF Gold P: 12,016 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.
HW Helper
Thanks
P: 10,540
 Quote by 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 http://="http://www.sosmath.com/trig...ig5.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 Va(t)=Asin(wt) and Vb(t)=Bsin(wt+β) as Vc(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 :

Vb= B [sin(wt)cos(β)+cos(wt)sin(β)] and Vc=C sin(wt+γ)=C[sin(wt)cos(γ)+cos(wt)sin(γ)].

With those, the equation Va+Vb = Vc 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
 P: 11 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.
 HW Helper Thanks P: 10,540 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
 P: 11 Thank you ehild
 Sci Advisor HW Helper PF Gold P: 12,016 Going the path along with the half angle product formula would have given you precisely the same expression.
 P: 11 Thank you arildno I understand that now. It took a while but I got there in the end.
P: 446
 Quote by arildno Going the path along with the half angle product formula would have given you precisely the same expression.
With a lot less work!
 HW Helper Thanks P: 10,540 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