Find Resultant Wave of y1 & y2: Get Hint Now!

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The discussion revolves around finding the resultant wave from two given waves, y1=3sin(kx-wt) and y2=4cos(kx+wt). A participant initially proposed the resultant wave as y= 7 cos(wt+45) sin(kx+45) but was questioned about its accuracy. Other contributors provided insights on using trigonometric identities to combine the waves, emphasizing the importance of correctly applying the addition formulas for sine and cosine. It was noted that the arguments for the sine and cosine functions differ, complicating the addition process. Ultimately, the consensus is that the initial answer is incorrect, and participants encouraged further exploration using the provided hints.
moham_87
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Hiiiiiiiiiii everyone
I've these two waves
y1=3sin(kx-wt)
y2=4cos(kx+wt)
I need to find the resultant wave (y1+y2)
I got that answer:

y= 7 cos(wt+45) sin(kx+45)
is that right?? please if not give me a hint

==================My efforts=============================
I added the amplidtude mathematically
then using trigonometric rules i added the two "sin" and "cos" functions
but i still need to know if my answer is true

bYYyYyyyYYyYYyYE
 
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The addition formula that I have for the sum of waves with different amplitudes is:

A cos(x) + B sin(x) = \sqrt {A^2 + B^2} cos(x \pm \delta)
tan( \delta) = \frac {sin \delta} {cos \delta} = \pm \frac B A
 
Integral said:
The addition formula that I have for the sum of waves with different amplitudes is:

A cos(x) + B sin(x) = \sqrt {A^2 + B^2} cos(x \pm \delta)
tan( \delta) = \frac {sin \delta} {cos \delta} = \pm \frac B A

What is that segma?
 
and what about if...

What about if "x" is not equal in both equations??
 
Integral,

The arguments for the sin and cos in your identity are both x. I may be missing a simplification you're seeing, but he's got one wave going left and one going right, so his arguments aren't the same.

I tried using the identities for the sum and difference of two angles, but nothing cancelled, so it just got messy. But like I said, I may be missing something.
 
yeah it gets a little messy but:

3 \sin {(kx-\omega t)}= 3( \sin{kx} \cos{\omega t }- \sin{\omega t} \cos {kx} )
and
4\cos{(kx+\omega t)}= 4(\cos{kx} \cos {\omega t}+ \sin{kx} \sin{\omega t})
add these together to get

\cos {kx} (4 \cos {\omega t} - 3 \sin {\omega t})+ \sin{kx}(3 \cos {\omega t}+ 4 \sin{\omega t})

now apply the formula in my first post to the terms in parentheses.

the \delta (thats a low case delta) is defined in the second line of my first post.
 
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Integral, nicely done!

Ok moham87, it doesn't look like your answer's right. But you've got your hint, so have at it! And watch those signs, or they'll kill you.
 
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