Precalc help please!?

Question

if sin t = 4/5 and cos t < 0, then find sin(2t).

Answer 1

sin t = 4/5 means that the ratio of the opposite to the hypotenuse is 4/5. So the adjacent side is going to be +/- sqrt(5^2 - 4^2) = +/-3. Since you're given that cos t < 0, the adjacent must be -3, so that cos t = -3/5.

Then sin 2t = 2sin t cos t = 2(4/5)(-3/5) = -24/25

Answer 2

sin t = 4/5, so 0 < t < 90 or 90 < t 180, sin is negative if t > 180
but cos t < 0, then 90 < t 180 or 270 < t < 360

thus 90 < t 180
since sin t = 4/5, a closer limit would be 120 < 135

the rest is geometry

Answer 3

Using the formula (sin t)^2 + (cos t)^2 = 1, we find that (cos t)^2 = 9 / 25 thus cos t = -3 / 5. Then sin(2t) = 2 (sin t) (cos t) = - 24 / 25.