In the following laplace transform, how do you get rid of the cos(2at)*sin(at)?

l{sin^3 (at)}
l{ sin(at) * [1/2 ( 1 - cos(2at)]}
1/2 l{sin(at)} - 1/2 l{cos(2at)*sin(at)}
1/2 * (a^2)/(s^2 + a^2) - 1/2 l{cos(2at)*sin(at)}

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

3; 9 is 8 to -4

step-by-step explanation: