Adding two sinusoidal waves of same frequency but out of phase

[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)

 

[anuttarasammyak]

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

 

[souky101]

.
Thanks for your reply
But how you can drive it mathematically
.

 

[DaveE]

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 $\frac{D}{C} = \frac{Esin(\theta)}{Ecos(\theta)} = tan{\theta}$

 

[anuttarasammyak]

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
$$F=C \cos \phi, L=C\sin \phi$$