Applying trigonometric functions: time particle at (0, 0)

[mcof]

Hi,

A particle is moving in a plane. The x and y coordinates of the particle are given by x = sin48t and y = cos25t, where t is time. A time t = 0, the particle is at the point (0,1). What are the values of t for which the particle is at the origin.

Can I just ask how to find the time (t)?
I thought that t = sin x/48 or t = cos y/25, but it will all equal to zero. (t = sin 0/48)
Is there a formula for time (t)? Thank you so much for the help!

mcof

 

[stapel]

The x and y coordinates of the particle are given by x = sin48t and y = cos25t, where t is time. A time t = 0, the particle is at the point (0,1). What are the values of t for which the particle is at the origin.

For what values of t is sin(48t) equal to zero?

For what values of t is cos(25t) equal to zero?

For which of those values are both equal to zero?

Eliz.

 

[Deleted member 4993]

Duplicate post:

http://www.freemathhelp.com/forum/viewt ... 10&t=28456

Looks like you did not solve the problem \.

for t = (n/48)*(pi) --> we have x = 0

for t = [(2m+1)/50] * (pi) --> we have y = 0

for what 'integer values' of 'm' and 'n' are those two 't' equal

When n = 24 and m = 12 we have t = pi/2 and

x = sin (48 t) = sin (24 pi) = 0

y = cos (25 t) = cos (25/2 pi) = 0