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Trig Form Conversion question

  • Thread starter Sparky_
  • Start date
  • #1
196
1

Homework Statement



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

[tex] -cos(\sqrt(2)t) - 2sin(\sqrt(2)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^{-1}phi = \frac{c1}{A} [/tex]
[tex] phi = cos^{-1} phi = \frac{c2}{A} [/tex]
and
[tex] phi = tan^{-1} phi = \frac{c1}{c2} [/tex]


The Attempt at a Solution




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]

and
[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.

Thanks
-Sparky
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
A = 5
A isn't 5. You miswrote your answer here, despite having found the correct one.

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

[tex]sin \ \phi = -\frac{1}{\sqrt{5}}[/tex] 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 [tex]\pi + \phi \ \mbox{or} \ -\phi[/tex]'

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

The only possible solution which satisfies both is [tex]\pi + \phi[/tex].
 
  • #3
196
1
A isn't 5. You miswrote your answer here, despite having found the correct one.
Oops, typing error, I meant [tex] A= \sqrt(-1^2 + -2^2) [/tex]

[tex] A= \sqrt(5) [/tex]

Thanks
 

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