Examine linear dependence of f_1(x)=1, f_2(x)=sin^2x…

Examine linear dependence of $f_1(x)=1,\qquad f_2(x)=\sin^2x,\qquad f_3(x)=\cos2x$.

Step 1: Direct check via the double-angle identity

Recall $\cos2x=1-2\sin^2x$. Rearranged: \[1\cdot1+(-2)\cdot\sin^2x+(-1)\cdot\cos2x =(1-2\sin^2x)-\cos2x=\cos2x-\cos2x=0.\]

This holds for every $x$, with constants $c_1=1,\ c_2=-2,\ c_3=-1$, not all zero.

Step 2: Wronskian confirmation

Derivatives: $f_1'=f_1''=0$; $f_2'=\sin2x,\ f_2''=2\cos2x$; $f_3'=-2\sin2x,\ f_3''=-4\cos2x$. Expanding along the first column (only the top entry survives): \[W(x)=\begin{vmatrix}1&\sin^2x&\cos2x\\0&\sin2x&-2\sin2x\\ 0&2\cos2x&-4\cos2x\end{vmatrix} =\begin{vmatrix}\sin2x&-2\sin2x\\2\cos2x&-4\cos2x\end{vmatrix}\]

\[ =-4\sin2x\cos2x+4\sin2x\cos2x=0. \]

$W\equiv0$, consistent with (and here, together with an explicit relation, confirming) linear dependence.

Step 3: Conclusion

\[1-2\sin^2x-\cos2x\equiv0\quad\Longrightarrow\quad \cos2x=1-2\sin^2x\]

The system $\{1,\ \sin^2x,\ \cos2x\}$ is linearly dependent: $\cos2x$ is a linear combination of $1$ and $\sin^2x$.