Determine whether the functions \( \mathrm{y}_{1} \) and \( \mathrm{y}_{2} \) are linearly dependent on the interval \( (0,1) \). \[ y_{1}=1-2 \sin ^{2} t \cdot y_{2}=6 \cos 2 t \] Select the correct choice below and, if necessary, fill in the answer box within your choice. A. Since \( y_{1}= \) \( \square \) \( y_{2} \) on \( (0,1) \), the functions are linearly independent on \( (0,1) \). (Simplify your answer.) B. Since \( y_{1}=\left(\square y_{2}\right. \) on \( (0,1) \), the functions are linearly dependent on \( (0,1) \). (Simplify your answer.) \( \square \) (Simplify your answer.) C. Since \( y_{1} \) is not a constant multiple of \( y_{2} \) on ( 0,1 ). the functions are linearly independent on ( 0,1 ). D. Since \( y_{1} \) is not a constant multiple of \( y_{2} \) on ( 0,1 ). the functions are linearly dependent on \( (0,1) \).
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To determine whether the functions \( y_{1} = 1 - 2 \sin^2 t \) and \( y_{2} = 6 \cos 2t \) are linearly dependent or independent, we can express \( y_1 \) in terms of \( y_2 \). We know that \( \cos 2t = 1 - 2 \sin^2 t \), so we can rewrite \( y_{1} \) as \( y_{1} = 1 - 2 \sin^2 t = \cos 2t \). Thus, \( y_{1} = \frac{1}{6} y_{2} \) on \( (0,1) \), making them linearly dependent. So, we can fill in the blanks as follows: B. Since \( y_{1}=\left(\frac{1}{6} y_{2}\right) \) on \( (0,1) \), the functions are linearly dependent on \( (0,1) \). \( \frac{1}{6} \)
