Answer :
The functions f(x) = sin(a) and g(x) = cos(a) are shown to be linearly independent using the Wronskian. Additionally, for a ∈ R, the functions f₁(x) = sin(x + a) and g₁(x) = cos(x + a) are shown to belong to the span of f and g. Finally, it is demonstrated that the set {f₁, g₁} is linearly independent.
To show that f(x) = sin(a) and g(x) = cos(a) are linearly independent, we can compute their Wronskian. The Wronskian of two functions is defined as the determinant of the matrix [f, g]' = [f(x), g(x)]'. In this case, the Wronskian is:
W(f, g) = |sin(a) cos(a)|
|-cos(a) sin(a)|
Calculating the determinant, we get:
det(W(f, g)) = sin(a)*sin(a) - (-cos(a)*cos(a)) = sin^2(a) + cos^2(a) = 1
Since the Wronskian is nonzero (det(W(f, g)) ≠ 0), the functions f and g are linearly independent.
Next, we consider the functions f₁(x) = sin(x + a) and g₁(x) = cos(x + a) for a ∈ R. We want to show that these functions belong to the span of f(x) = sin(a) and g(x) = cos(a). By using trigonometric identities, we can express f₁ and g₁ in terms of f and g:
f₁(x) = sin(x + a) = sin(x)cos(a) + cos(x)sin(a) = cos(a)*f(x) + sin(a)*g(x)
g₁(x) = cos(x + a) = cos(x)cos(a) - sin(x)sin(a) = cos(a)*g(x) - sin(a)*f(x)
This shows that both f₁(x) and g₁(x) can be written as linear combinations of f(x) and g(x), implying that f₁ and g₁ belong to the span of f and g.
Finally, to show that {f₁, g₁} is linearly independent, we can again calculate the Wronskian:
W(f₁, g₁) = |cos(a)*f(x) + sin(a)*g(x) cos(a)*g(x) - sin(a)*f(x)|
|cos(a)*f'(x) + sin(a)*g'(x) cos(a)*g'(x) - sin(a)*f'(x)|
Expanding the determinant, we find that det(W(f₁, g₁)) = sin^2(a) + cos^2(a) = 1. Since the Wronskian is nonzero, it indicates that {f₁, g₁} is linearly independent.
In conclusion, we have shown that f(x) = sin(a) and g(x) = cos(a) are linearly independent using the Wronskian. Additionally, for a ∈ R, f₁(x) = sin(x + a) and g₁(x) = cos(x + a) belong to the span of f and g, and the set {f₁, g₁} is linearly independent.
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