# Adding two sinusoidal waves of same frequency but out of phase

• souky101

#### souky101

Homework Statement
Adding two sinusoidal waves of same frequency but out of phase - Does the resultant wave form still pure sinusoidal with same frequency?
Relevant Equations
Asin(wt) + Bsin(wt+a)
Asin(wt)}+Bsin(wt+a)
Asin(wt) +B sin(wt)cos(a) +Bcos(wt)sin(a)
Asin(wt) + ksin(wt) + Lcos( wt)
(A+K) sin(wt) + Lcos(wt)
Fsin(wt) + Lcos(wt)

Delta2

It can be deduced to the form of
$$C\sin(\omega t+\phi)$$

Delta2
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But how you can drive it mathematically
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You started out OK. It's mostly based on the trig identity you used ##sin(a+b)=sin(a)cos(b) + cos(a)sin(b)##
So:

##Asin(\omega t) + B sin(\omega t + \phi) =##
##Asin(\omega t)+B(sin(\omega t)cos(\phi)+cos(\omega t)sin(\phi)) =##
##(A+Bcos(\phi))sin(\omega t)+Bsin(\phi)cos(\omega t) \equiv Csin(\omega t)+Dcos(\omega t) ##
where ##C \equiv A+Bcos(\phi)## and ##D \equiv Bsin(\phi)##
since sin(x) and cos(x) are orthogonal, we can used Pythagoras to simplify to
##Asin(\omega t) + B sin(\omega t + \phi) = \sqrt{C^2+D^2}sin(\omega t + \theta)##
where ##\theta = tan^{-1}(\frac{D}{C})##

The last bit comes from working that original identity in reverse:
Assume the answer has the form ##Csin(\omega t)+Dcos(\omega t) \equiv Esin(\omega t + \theta)##
Then ##Esin(\omega t + \theta) = E cos(\theta)sin(\omega t)+Esin(\theta)cos(\omega t)## from the identity.
This means ##C \equiv Ecos(\theta)## and ##D \equiv Esin(\theta)##
Then note that ## C^2+D^2=E^2cos^2(\theta)+E^2sin^2(\theta)= E^2## so ##E=\sqrt{C^2+D^2}##
Also ##\frac{D}{C} = \frac{Esin(\theta)}{Ecos(\theta)} = tan{\theta}##

FactChecker and Delta2
But how you can drive it mathematically
Let us equate them as
$$F \sin \omega t + L \cos\omega t = C \sin (\omega t + \phi)$$
Expanding RHS you see
$$F=C \cos \phi, L=C\sin \phi$$

Delta2