The motion of an oscillator is described by the equation x(t)=sin t+sin2t. What is x if cos t=-0.25 and t is in the interval (180, 270)?

First of all, before calculate sin t, we must establish to
what quadrant belongs. Due to the facts from hypothesis, t is in the interval (pi,
3pi/2), so the angle t belongs to the third quadrant, where the value of the function
sine is negative.

cos a = -.25
= -1/4

sin a  = sqrt[1- (-1/4) (from the fundamental
formula of trigonometry,where (sin a)^2 + (cosa)^2  =
1).

sin a = -sqrt(15)/4

To
determine x, first we have to calculate sin 2t.

We'll apply
the formula for the double angle:

sin 2a = sin
(a+a)=sina*cosa + sina*cosa=2sina*cosa

We'll substitute 2a
by 2t and we'll re-write the equation x(t).

x(t) = sin t +
2sint*cos t

e= -sqrt(15)/4 +
2*(1/4)*sqrt(15)/4

We'll calculate the LCD of the
ratios:

LCD = 16

We'll
factorize by sqrt(15)/4:

x(t) = [sqrt(15)/4](-1 +
2/4)

x(t) = [sqrt(15)/4](-1 +
1/2)

x(t) =
[-sqrt(15)/8]