# Trig Form Question

## Homework Statement

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

$$-cos(sqrt(2)t) - 2sin(sqrt(s)t)$$

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

$$Asin(wt + phi)$$

## Homework Equations

$$A= sqrt(-1^2 + -2^2)$$

$$phi = sin^-1phi = \frac{c1}{A}$$
$$phi = cos^-1 phi = \frac{c2}{A}$$
and
$$phi = tan^-1 phi = \frac{c1}{c2}$$

## The Attempt at a Solution

A = 5

$$phi = sin^-1phi = \frac{-1}{sqrt(5)}$$
$$phi =$$

$$phi = cos^-1 phi = \frac{-2}{sqrt(5)}$$
$$phi =$$

and
$$phi = tan^-1 phi = \frac{-1}{-2}$$
$$phi =$$

## Answers and Replies

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CompuChip
Science Advisor
Homework Helper
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 ($\omega = \sqrt{2}$). Then the amplitude is indeed $A = \sqrt{5}$ (you gave the correct formula, but forgot the root in the later result. Also watch out to write $(-1)^2 = (-1) \times (-1) = 1$, which is something different than $-1^2 = -(1 \times 1) = -1$). Then just plug in t = 0 and you get an equation for phi, which you can solve, giving $\phi = \sin^{-1}(1/\sqrt{5}) + \pi$ (if you want a value between 0 and 2 pi).

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

$$phi = sin^{-1}phi = \frac{-1}{\sqrt(5)}$$
$$phi = -0.4636$$

$$phi = cos^{-1} phi = \frac{-2}{\sqrt(5)}$$
$$phi = 2.678$$

and
$$phi = tan^{-1} phi = \frac{-1}{-2}$$
$$phi = 0.4636$$

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.

Thanks
-Sparky

Last edited:
CompuChip
Science Advisor
Homework Helper
So you have found A and omega, and now you want
$$A \sin(\omega t + \phi) = - \cos(\sqrt{2} t) - 2 \sin(\sqrt{2} t)$$.
If you look at the point t = 0, you get the equation
$$A \sin(\phi) = - 1$$
from which you can solve $\phi$. So indeed, you get the equation
$$\phi = \operatorname{arcsin}(- 1 / \sqrt{5})$$
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.