DrewD said:
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
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 :
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
2cos
2(γ)=C
2
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