Please Help! Pre-Calculus and Trigonometry.

Question

Give the exact values of sin2x, cos2x, and tan2x.

sin x = 3/5, pie/2 ≤ x < pie

Answer 1

sin2x = 2 sinx cos x

cos2x = 1 - 2 (sin x) **2

tan2x = sin 2x / cos 2x

We are given sin x = 3/5.

Because pi/2 < x < pi, x is located in the 2nd quadrant, where sin x is positive and cos x is negative.

SInce (sin x)**2 + (cos x)**2 = 1

(3/5)**2 + (cos x)**2 = 1

9/25 + (cos x)**2 = 1

(cos x)**2 = 16/25

cos x = +/- 4/5, but since in 2nd quadrant, cos x = -4/5

tan x = sin x/cos x = 3/5 / - 4/5 = -3/4

Plugging those values into our formulas:

sin2x = 2 sinx cos x = 2 * 3/5 * -4/5 = -24/25

cos2x = 1 - 2 (sin x) **2 = 1 - 2 (3/5)**2 = 7/25

tan2x = sin 2x / cos 2x = (-24/25)/(7/25) = -24/7