Adding Sine and Cosine Waves- How to get formula

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 2K views
opus
Gold Member
Messages
717
Reaction score
131
I have included a screenshot of a part of my textbook that is giving me a slight bit of confusion.

It's talking about how to get the formula for adding sines and cosines.

The part that I am confused about is the very first formula introduced in the screenshot.

From what I understand, we are taking one side of the sum of sines formula, and Asin(x)+Bcos(x).
The part in parentheses I do understand. It's stating the sum of sines formula in a different way. But I do not understand why the ##\sqrt{A^2+B^2}##, which would be the hypotenuse, is on the outside of the parentheses.
 

Attachments

  • Screen Shot 2018-06-08 at 7.19.33 PM.png
    Screen Shot 2018-06-08 at 7.19.33 PM.png
    33.6 KB · Views: 529
Physics news on Phys.org
Here’s a shot of my understanding of it thus far.
 

Attachments

  • ACE46C74-496B-4CB6-A4E2-0605D26D14F7.png
    ACE46C74-496B-4CB6-A4E2-0605D26D14F7.png
    13.4 KB · Views: 516
opus said:
I have included a screenshot of a part of my textbook that is giving me a slight bit of confusion.

It's talking about how to get the formula for adding sines and cosines.

The part that I am confused about is the very first formula introduced in the screenshot.

From what I understand, we are taking one side of the sum of sines formula, and Asin(x)+Bcos(x).
The part in parentheses I do understand. It's stating the sum of sines formula in a different way. But I do not understand why the ##\sqrt{A^2+B^2}##, which would be the hypotenuse, is on the outside of the parentheses.
In a thread earlier today, you learned that you could multiply one side of an equation by 1 and still maintain equality. So multiply ##A~sin(x) + B~cos(x)## by ##\frac {\sqrt{A^2+B^2}} {\sqrt{A^2+B^2}}## and see what you get. It should look a lot like the first line of the derivation.
 
  • Like
Likes   Reactions: opus
AH! That's the one! Threw me off a little because it looks like they did that, and they factored the square root term out of the numerator and left it in the denominator. That is perfect, thank you!
 
As an additional question, now that we have the Sum of Sines and Cosines formula
##Asin\left(x\right)+Bcos\left(x\right)## = ##\sqrt{A^2+B^2}sin\left(x+θ\right)##,
is the +θ considered a phase shift? That is, are we taking the graph of sin(x) and shifting it θ to the left?
 
  • Like
Likes   Reactions: tnich
opus said:
As an additional question, now that we have the Sum of Sines and Cosines formula
##Asin\left(x\right)+Bcos\left(x\right)## = ##\sqrt{A^2+B^2}sin\left(x+θ\right)##,
is the +θ considered a phase shift? That is, are we taking the graph of sin(x) and shifting it θ to the left?
Yes, that is exactly the case.
 
  • Like
Likes   Reactions: jedishrfu and opus
Thank you tnich.
 
  • Like
Likes   Reactions: jedishrfu