Determine whether the functions y 1 and y 2 are linearly dependent on the interval (0,1). y 1equals2 cosine squared t minus 1 , y 2equals6 cosine 2 t Select the correct choice below and, if necessary, fill in the answer box within your choice. A. Since y 1equals (nothing) y 2 on (0,1), the functions are linearly independent on (0,1). (Simplify your answer.) B. Since y 1equals (nothing) y 2 on (0,1), the functions are linearly dependent on (0,1). (Simplify your answer.) C. Since y 1 is not a constant multiple of y 2 on (0,1), the functions are linearly dependent on (0,1). D. Since y 1 is not a constant multiple of y 2 on (0,1), the functions are linearly independent on (0,1).

Question

Kristen Mccullough

Mathematics

5 March, 23:30

Determine whether the functions y 1 and y 2 are linearly dependent on the interval (0,1). y 1equals2 cosine squared t minus 1 , y 2equals6 cosine 2 t Select the correct choice below and, if necessary, fill in the answer box within your choice. A. Since y 1equals (nothing) y 2 on (0,1), the functions are linearly independent on (0,1). (Simplify your answer.) B. Since y 1equals (nothing) y 2 on (0,1), the functions are linearly dependent on (0,1). (Simplify your answer.) C. Since y 1 is not a constant multiple of y 2 on (0,1), the functions are linearly dependent on (0,1). D. Since y 1 is not a constant multiple of y 2 on (0,1), the functions are linearly independent on (0,1).

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Answers (1)

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Tripp Powell · Answer 1

D. Since y₁ is not a constant multiple of y₂ on (0,1), the functions are linearly independent on (0,1).

Step-by-step explanation:

y₁=2Cos²t-1

y₂=6Cos2t

We want to determine if y₁ and y₂ are linearly independent in the interval (0,1).

To do this, we show that there does not exist any c₁ and c₂ in (0,1) that makes the expression:

f (t) = c₁ (2Cos²t-1) + c₂6Cos2t=0.

If c₁ and c₂=0

f (t) = 0

Let c₁=1 and c₂=-⅓

f (t) = c₁ (2Cos²t-1) + c₂6Cos2t

=2Cos²t-1-⅓*6 (cos²t-sin²t)

=2Cos²t-1-2cos²t+2sin²t

=2sin²t-1

Since f (t) ≠0, y₁ is not a constant multiple of y₂ and the functions are linearly independent on (0,1).