If cos u = 4/5 and tan v = - 12/5 while (3pi)/2 < u < 2pi and pi/2 < v < pi , find : sin(u-v)

Expand sin(u - v) using the angle sum formula for sine:

sin(u - v) = sin(u) cos(v) - cos(u) sin(v)

Recall the Pythagorean identity: for all x,

cos²(x) + sin²(x) = 1

Dividing both sides by cos²(x) gives

1 + tan²(x) = sec²(x)

(since tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x))

With 3π/2 < u < 2π, we expect sin(u) < 0, and with π/2 < v < π, we expect cos(v) < 0 and sin(v) > 0.

Solve for sin(u) :

sin²(u) = 1 - (4/5)²

sin(u) = - √(1 - 16/25)

sin(u) = - √(9/25)

sin(u) = -3/5

Solve for sec(v) :

sec²(v) = 1 + (-12/5)²

sec(v) = - √(169/25)

sec(v) = -13/5

Take the reciprocal of both sides to solve for cos(v) :

1/sec(v) = 1 / (-13/5)

cos(v) = -5/13

Solve for sin(v) :

sin²(v) = 1 - cos²(v)

sin(v) = √(1 - (-5/13)²)

sin(v) = √(144/169)

sin(v) = 12/13

Now put everything together:

sin(u - v) = (-3/5) (-5/13) - (4/5) (12/13) = -33/65

c

-by-step explanation:

it would be c