Answer :
10.1 f and g are linearly independent using the Wronskian
10.2 f₁ and g₁ belong to W
10.3 {f₁,g₁} is also linearly independent.
10.1 To show that f and g are linearly independent, we can calculate their Wronskian, which is the determinant of the matrix [f g; f' g'], where f' and g' represent the derivatives of f and g with respect to x. If the Wronskian is non-zero for all x, then f and g are linearly independent. In this case, the Wronskian of f and g is sin(x)cos(x) - cos(x)sin(x) = 0. Since the Wronskian is identically zero, f and g are linearly independent.
10.2 To show that f₁ and g₁ belong to W, we need to demonstrate that they can be expressed as linear combinations of f and g. Using the trigonometric identities, we have f₁(x) = sin(x + a) = sin(x)cos(a) + cos(x)sin(a), and g₁(x) = cos(x + a) = cos(x)cos(a) - sin(x)sin(a). These expressions show that f₁ and g₁ are linear combinations of f and g, with coefficients cos(a) and sin(a).
10.3 To prove that {f₁,g₁} is linearly independent, we can use the Wronskian again. Calculate the Wronskian of f₁ and g₁, which is (cos(x)cos(a) - sin(x)sin(a))(cos(x) - sin(a)) - (sin(x)cos(a) + cos(x)sin(a))(sin(x) - cos(a)). Simplifying this expression, we get cos²(x) - sin²(x) - cos²(a) + sin²(a), which is non-zero for all x. Therefore, the Wronskian is non-zero, indicating that f₁ and g₁ are linearly independent.
Learn more about trigonometric identities here:
https://brainly.com/question/24377281
#SPJ11
