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