Final answer:
To find sin2x, cos2x, and tan2x when sinx is -1/√10 in quadrant III, use the double-angle formulas for sine and cosine, remembering the signs of functions in quadrant III. Calculations show sin2x = 1.8, cos2x = 0.8, and tan2x = 2.25.
Step-by-step explanation:
To find sin2x, cos2x, and tan2x given that sinx = -1/√10 and the angle x terminates in quadrant III, we use the double-angle formulas for sine and cosine, and then find the tangent from these results.
Double-Angle Formulas:
- sin 2x = 2 sin x cos x
- cos 2x can be computed using any of these:
- tan 2x = sin 2x / cos 2x
Calculation for Quadrant III:
In quadrant III, both sine and cosine are negative. Thus, if sinx = -1/√10, we find cosx using the Pythagorean identity sin²x + cos²x = 1, which implies cos²x = 1 - sin²x. Here, cos²x = 1 - (-1/√10)² = 1 - 1/10 = 9/10, and since cosx is negative in quadrant III, cosx = -√(9/10).
Now apply the double-angle formulas:
sin 2x = 2 * (-1/√10) * (-√(9/10)) = 2/√10 * √(9/10) = 18/10 or sin 2x = 1.8
cos 2x = 1 - 2*(-1√10)² = 1 - 2/10 = 8/10 or cos 2x = 0.8
tan 2x = sin 2x / cos 2x = 1.8 / 0.8 = 2.25