Examine linear dependence of f_1(x)=sin x, f_2(x)=cos x…

Examine linear dependence of $f_1(x)=\sin x,\qquad f_2(x)=\cos x,\qquad f_3(x)=\sin2x$.

Step 1: Direct check

Suppose $c_1\sin x+c_2\cos x+c_3\sin2x\equiv0$. Setting $x=0$ gives $c_2=0$; setting $x=\pi/2$ gives $c_1=0$. With $c_1=c_2=0$, the identity reduces to $c_3\sin2x\equiv0$, and since $\sin2x\not\equiv0$, also $c_3=0$.

\[ c_1=c_2=c_3=0. \]

Only the trivial combination vanishes identically, even though $\sin2x=2\sin x\cos x$ is built from $f_1$ and $f_2$ – as a product, not a linear combination.

Step 2: Wronskian

Derivatives: $f_1'=\cos x,\ f_1''=-\sin x$; $f_2'=-\sin x,\ f_2''=-\cos x$; $f_3'=2\cos2x,\ f_3''=-4\sin2x$.

\[ W(x)=\begin{vmatrix}\sin x&\cos x&\sin2x\\ \cos x&-\sin x&2\cos2x\\ -\sin x&-\cos x&-4\sin2x\end{vmatrix} \]

Expanding along the first row, the three $2\times2$ minors are: \[\begin{vmatrix}-\sin x&2\cos2x\\-\cos x&-4\sin2x\end{vmatrix}=4\sin x\sin2x+2\cos x\cos2x,\qquad \begin{vmatrix}\cos x&2\cos2x\\-\sin x&-4\sin2x\end{vmatrix}=-4\cos x\sin2x+2\sin x\cos2x,\]

\[ \begin{vmatrix}\cos x&-\sin x\\-\sin x&-\cos x\end{vmatrix}=-\cos^2x-\sin^2x=-1. \]

Substituting: $W=\sin x(4\sin x\sin2x+2\cos x\cos2x)-\cos x(-4\cos x\sin2x+2\sin x\cos2x)+\sin2x(-1)$.

\[ =4\sin2x(\sin^2x+\cos^2x)+\underbrace{2\sin x\cos x\cos2x-2\sin x\cos x\cos2x}_{=\,0}-\sin2x =4\sin2x-\sin2x. \] $W(x)=3\sin2x$.

Since $W(\pi/4)=3\neq0$, in particular $W\not\equiv0$.

Step 3: Conclusion

Both tests agree: the system $\{\sin x,\ \cos x,\ \sin2x\}$ is linearly independent. The relation $\sin2x=2\sin x\cos x$ is a genuine (nonlinear) product identity, not a counterexample to independence.