The complete set of real values of 'a' such that the point lies triangle `p(a, sin a)` lies inside the triangle formed by the lines `x-2y +2= 0; x+y=0` and `x-y-pi=0`

Explanation

To solve the problem of finding the complete set of real values of 'a' such that the point \( P(a, \sin a) \) lies inside the triangle formed by the lines \( x - 2y + 2 = 0 \), \( x + y = 0 \), and \( x - y - \pi = 0 \), we will follow these steps:

### Step 1: Find the vertices of the triangle formed by the lines.

1. **Find the intersection of the lines \( x - 2y + 2 = 0 \) and \( x + y = 0 \)**:
- From \( x + y = 0 \), we have \( y = -x \).
- Substitute \( y = -x \) into \( x - 2(-x) + 2 = 0 \):
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