Answer :
10.1 By calculating the Wronskian of f(x) = sin(x) and g(x) = cos(x), we can show that f and g are linearly independent.
10.2 By evaluating f₁(x) = sin(x + a) and g₁(x) = cos(x + a) and checking if they belong to the span W, we can demonstrate that f₁ and g₁ are in W.
10.3 To establish that {f₁, g₁} is linearly independent, we need to show that no nontrivial linear combination of f₁ and g₁ can equal the zero function.
10.1 To prove that f(x) = sin(x) and g(x) = cos(x) are linearly independent, we calculate the Wronskian:
W(f, g) = det | f g |
| f' g' |
By evaluating the determinant with f(x) = sin(x) and g(x) = cos(x), we find that W(f, g) = sin(x)cos(x) - cos(x)sin(x) = 0. Since the Wronskian is zero for all x, f and g are linearly independent.
10.2 For f₁(x) = sin(x + a) and g₁(x) = cos(x + a), we substitute them into the expressions and observe that f₁(x) = sin(x)cos(a) + cos(x)sin(a) and g₁(x) = cos(x)cos(a) - sin(x)sin(a). We can rewrite f₁(x) and g₁(x) as linear combinations of f(x) = sin(x) and g(x) = cos(x) by using trigonometric identities. Therefore, f₁ and g₁ belong to the span W.
10.3 To show that {f₁, g₁} is linearly independent, we assume a nontrivial linear combination of f₁ and g₁, c₁f₁ + c₂g₁, equals the zero function. By equating coefficients and using trigonometric identities, we can deduce that c₁ = c₂ = 0 is the only solution. Thus, {f₁, g₁} is linearly independent.
Learn more about trigonometric identities here: brainly.com/question/24377281
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