Finding the Phase Angle for a Differential Equation Solution

[Sparky_]

Homework Statement

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

-cos(\sqrt(2)t) - 2sin(\sqrt(2)t)

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

Asin(wt + phi)

Homework Equations

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

phi = sin^{-1}phi = \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^{-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

 

[Defennder]

A = 5

A isn't 5. You miswrote your answer here, despite having found the correct one.

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

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

The "basic angle" (do you use that term? I don't know if this is the proper term) of \phi is 0.4636.

sin \ \phi = -\frac{1}{\sqrt{5}} Are you familiar with the trigonometric quadrants ASTC, where A is the top right hand quadrant and the rest are labeled in that order anti-clockwise?

From that diagram, sin is negative for the bottom 2 quadrants. That means to say the angle phi is either \pi + \phi \ \mbox{or} \ -\phi'

cos \ \phi = -\frac{2}{\sqrt{5}} Again you want the values of the angle for which cos phi is negative. From the ASTC quadrant diagram, cosine is negative for \pi - \phi \ \mbox{or} \ \pi + \phi

The only possible solution which satisfies both is \pi + \phi.

 

[Sparky_]

A isn't 5. You miswrote your answer here, despite having found the correct one.

Oops, typing error, I meant A= \sqrt(-1^2 + -2^2)

A= \sqrt(5)

Thanks