- #1

- 734

- 0

## Main Question or Discussion Point

why can y(t) = A cos[omega(t)]+bsin[omega(t)]

equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and

tan(sigma)=B/A?

equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and

tan(sigma)=B/A?

- Thread starter asdf1
- Start date

- #1

- 734

- 0

why can y(t) = A cos[omega(t)]+bsin[omega(t)]

equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and

tan(sigma)=B/A?

equal Ccos[omega(t)-sigma], where C=(A^2 + B^2)^(1/2) and

tan(sigma)=B/A?

- #2

James R

Science Advisor

Homework Helper

Gold Member

- 600

- 15

[tex]\cos(a-b) = \cos a \cos b + \sin a \sin b[/tex]

Try expanding out [itex]C\cos(\omega t - \sigma)[/itex] using that formula, and see what you get.

- #3

- 734

- 0

i got C[cos(wt)cos(sigma)+sin(wt)sin(sigma)]...

am i missing a step?

@@a

am i missing a step?

@@a

- #4

James R

Science Advisor

Homework Helper

Gold Member

- 600

- 15

What do you get if you eliminate sigma from these two equations?

- #5

- 734

- 0

C[cos(wt)A/C + sin(wt)B/C]= the orignal equation!

thanks!

thanks!

- #6

James R

Science Advisor

Homework Helper

Gold Member

- 600

- 15

Note also that [itex]A^2 + B^2 = C^2[/itex] and tan(sigma) = B/A, with the given definitions.

- #7

- 734

- 0

cool! thanks again! :)

- Replies
- 6

- Views
- 514

- Last Post

- Replies
- 2

- Views
- 620

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 536

- Last Post

- Replies
- 1

- Views
- 11K

- Last Post

- Replies
- 1

- Views
- 764

- Last Post

- Replies
- 5

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 32K

- Last Post

- Replies
- 69

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 2K