Simplify the equation - question about the angle

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Homework Help Overview

The discussion revolves around simplifying an equation involving trigonometric functions and exponential decay. The original poster seeks clarification on how a specific angle, φ = π/3, is derived from the equation x(t) = 5e(-t/5) (√3 sin(t) + cos(t)).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between coefficients of sine and cosine functions and their corresponding angles. There is an attempt to rewrite the equation in a form that highlights the angle φ.

Discussion Status

Some participants have offered insights into calculating φ based on the coefficients of the sine and cosine terms. There is recognition of the need for practice in understanding these concepts, and a supportive atmosphere for learning is evident.

Contextual Notes

There are indications that the original poster may have encountered similar concepts in previous discussions, suggesting a learning curve associated with these mathematical principles.

Vital
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Homework Statement


Hello!
Please, take a look at the following equation and help me to understand where the authors got the value of φ = π/3. I don't see where it is derived from as no additional conditions are given.

Homework Equations


x(t) = 5e(-t/5) cos(t) + 5e(-t/5) √3 sin(t)

The Attempt at a Solution



Here is how I proceed:
x(t) = 5e(-t/5) (cos(t) + √3 sin(t))

Now, given the formula:
f(x) = a sin(wx) + b cos(wx) + B (w > 0) is the same as
f(x) = √a2+b2 sin(wx + φ) + B

In my case, B = 0, so I rewrite the expression as:
x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )

x(t) = 5e(-t/5) √√32+12 sin(t + φ)

x(t) = 10e(-t/5) sin(t + φ)

But in the book they have:
x(t) = 10e(-t/5) sin(t + π/3)

Where did they get φ = π/3 from?

Thank you!
 
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x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )=##10e^{-t/5}\left(\sin {t }\left(\frac{\sqrt{3}}{2}\right)+\cos{t}\left(\frac{1}{2}\right)\right)##

##\sin{(t+\phi)}=\sin t \cos{\phi}+\cos{t}\sin{\phi}##

What is the value of ##\phi##?
 
Chestermiller said:
x(t) = 5e(-t/5) ( √3 sin(t) + cos(t) )=##10e^{-t/5}\left(\sin {t }\left(\frac{\sqrt{3}}{2}\right)+\cos{t}\left(\frac{1}{2}\right)\right)##

##\sin{(t+\phi)}=\sin t \cos{\phi}+\cos{t}\sin{\phi}##

What is the value of ##\phi##?
Ah! I see. I should have computed the value of φ based on the coefficients' values:

coefficient1 x cos(φ) = √3
coefficient2 x sin(φ) = 1
(coefficient1 x cos(φ))2 + (coefficient2 x sin(φ))2 = coefficient2
hence, coefficient = 2, and thus cos(φ) = √3/2 and sin(φ) = ½, and of course φ = π/3
Now I see how the value appeared.
 
Vital said:
Ah! I see. I should have computed the value of φ based on the coefficients' values:

coefficient1 x cos(φ) = √3
coefficient2 x sin(φ) = 1
(coefficient1 x cos(φ))2 + (coefficient2 x sin(φ))2 = coefficient2
hence, coefficient = 2, and thus cos(φ) = √3/2 and sin(φ) = ½, and of course φ = π/3
Now I see how the value appeared.

You have been shown this more than once already in other threads. It is something you need to learn and commit to memory, especially if you are going into Phhsics or Electrical Engineering.
 
Ray Vickson said:
You have been shown this more than once already in other threads. It is something you need to learn and commit to memory, especially if you are going into Phhsics or Electrical Engineering.
It takes practice and time to remember these notions. I am doing my best, and as I am learning on my own I don't have enough practice. Happy to be able to ask here. :-)
 

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