Given that θ is a minimum internal angle in a triangle, and the square of a (COS θ / 2) + (sin θ / 2) + (COS θ / 2) - the square of a (sin θ / 2) = a + 1, then the value range of a is A、a-1 C、a≤-3 D、a≥-3

Given that θ is a minimum internal angle in a triangle, and the square of a (COS θ / 2) + (sin θ / 2) + (COS θ / 2) - the square of a (sin θ / 2) = a + 1, then the value range of a is A、a-1 C、a≤-3 D、a≥-3

Because a (COS θ / 2) & sup2; + (sin θ / 2) & sup2; - (COS θ / 2) & sup2; - A (sin θ / 2) & sup2; = a + 1
So: a [(COS θ / 2) & sup2; - (sin θ / 2) & sup2;] - [(COS θ / 2) & sup2; - (sin θ / 2) & sup2;] = a + 1
So: a cos θ - cos θ = a + 1
So: (A-1) cos θ = a + 1
Obviously, a ≠ 1
So: cos θ = (a + 1) / (A-1)
Because theta is the smallest internal angle in a triangle
Therefore: 0 °＜θ≤ 60 °
So: 1 / 2 ≤ cos θ < 1
So: 1 / 2 ＜ (a + 1) / (A-1) ＜ 1
The solution (a + 1) / (A-1) < 1, that is: (a + 1) / (A-1) - 1 < 0
So: 2 / (A-1) < 0
Therefore: a < 1
The solution (a + 1) / (A-1) ≥ 1 / 2, that is: (a + 1) / (A-1) - 1 / 2 ≥ 0
So: (a + 3) / [2 (A-1)] ≥ 0
So: a > 1 or a ≤ - 3
So: a ≤ - 3
Select C, a ≤ - 3