Question: I'm worried that this will be a spanner in the works when I'm using a calculator in an exam , I know that $-2^2= 4$ since $(-2) \cdot (-2) = 4 $. So why does my calculator show it as negative 4 ? Does anyone know if there is a way to reset my calculator so it doesn't do this? The make and model is Casio fx-83ES.

Answer: 
Since $x$, $e^x$, and $\sin x$ are analytic, then their non-vanishing [Wronskian](https://en.wikipedia.org/wiki/Wronskian#The_Wronskian_and_linear_independence) is sufficient to conclude that the set is linearly independent. 

The Wronskian is equal to the determinant $W$ of the matrix given by 

$$\begin{bmatrix}
x & e^x & \sin x\\
1 & e^x & \cos x\\
0 & e^x & -\sin x\\
\end{bmatrix}$$

Therefore, we have 

$$W=
\begin{vmatrix}
x & e^x & \sin x\\
1 & e^x & \cos x\\
0 & e^x & -\sin x\\
\end{vmatrix}
=-xe^x(\sin x+\cos x)+2e^x \sin x$$

Inasmuch as $W\ne 0$ for all $x\in \mathscr{R}$ (e.g., take $x=\pi/2$), then $x$, $e^x$ and $\sin x$ are linearly independent.
Suppose
$ax +be^x +c\sin(x) =0
$.

If
$b \ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.

Therefore
$b = 0$,
so we have
$ax+c\sin(x)
= 0
$.

Again,
if $a \ne 0$,
we can choose $x$
large enough so that
$|ax| > |c|$.
Since
$|\sin(x)| \le 1$,
this can not hold.

Therefore
$a = 0$,
so that
$c \sin(x) = 0$.

This obviously implies that
$c=0$.

And we are done.