Trig Form Question

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  • #1

Homework Statement

This is a solutuion to a differential equation I've solved:

[tex] -cos(sqrt(2)t) - 2sin(sqrt(s)t) [/tex]

I've been asked to put it in the form:

[tex] Asin(wt + phi) [/tex]

Homework Equations

[tex] A= sqrt(-1^2 + -2^2) [/tex]

[tex] phi = sin^-1phi = \frac{c1}{A} [/tex]
[tex] phi = cos^-1 phi = \frac{c2}{A} [/tex]
[tex] phi = tan^-1 phi = \frac{c1}{c2} [/tex]

The Attempt at a Solution

A = 5

[tex] phi = sin^-1phi = \frac{-1}{sqrt(5)} [/tex]
[tex] phi = [/tex]

[tex] phi = cos^-1 phi = \frac{-2}{sqrt(5)} [/tex]
[tex] phi = [/tex]

[tex] phi = tan^-1 phi = \frac{-1}{-2} [/tex]
[tex] phi = [/tex]
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  • #2
Note that if s inside the sine is not equal to 2, then you cannot write it as a single sine at all. So the frequencies have to be the same, and you can just read them off ([itex]\omega = \sqrt{2}[/itex]). Then the amplitude is indeed [itex]A = \sqrt{5}[/itex] (you gave the correct formula, but forgot the root in the later result. Also watch out to write [itex](-1)^2 = (-1) \times (-1) = 1[/itex], which is something different than [itex]-1^2 = -(1 \times 1) = -1[/itex]). Then just plug in t = 0 and you get an equation for phi, which you can solve, giving [itex]\phi = \sin^{-1}(1/\sqrt{5}) + \pi[/itex] (if you want a value between 0 and 2 pi).
  • #3
I did not mean to "post" it yet - I meant to hit "preview".

Here are some more details and clean-up and my official question:

A = 5

[tex] phi = sin^{-1}phi = \frac{-1}{\sqrt(5)} [/tex]
[tex] phi = -0.4636[/tex]

[tex] phi = cos^{-1} phi = \frac{-2}{\sqrt(5)} [/tex]
[tex] phi = 2.678[/tex]

[tex] phi = tan^{-1} phi = \frac{-1}{-2} [/tex]
[tex] phi = 0.4636[/tex]

Can you help with the phase angle? (3 different answers - none of which are correct)

The book has 3.6052 which I see is pi + 0.4636.

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  • #4
So you have found A and omega, and now you want
[tex]A \sin(\omega t + \phi) = - \cos(\sqrt{2} t) - 2 \sin(\sqrt{2} t)[/tex].
If you look at the point t = 0, you get the equation
[tex]A \sin(\phi) = - 1[/tex]
from which you can solve [itex]\phi[/itex]. So indeed, you get the equation
[tex]\phi = \operatorname{arcsin}(- 1 / \sqrt{5})[/tex]
which you can take to be either -0.4636, or 0.4636 + pi. By graphing both you see that the latter is correct, and it is also neater, to express phi between 0 and 2 pi.

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