Help solving this question on even and odd functions a) Suppose that $E(x)$ is an even function and that $O(x)$ is an odd function. Suppose furthermore
that $E(x) + O(x) = 0$. Show that for all $x$, $E(x) = 0$ and $O(x) = 0$.
b) Use part a) to show that if $A\sin(ax)+B\cos(bx) = 0$ for all $x$, where $A,B,a,b$ are fixed real numbers, then $B = 0$ and one of either $A$ or $a$ is also equal to $0$.
 A: (a) We have 
$$E(x)+O(x)=0\tag{$1$}$$
for all $x$.
It follows that $E(-x)+O(-x)=0$ for all $x$.
But $E(-x)=E(x)$ and $O(-x)=-O(x)$. Thus 
$$E(x)-O(x)=0\tag{$2$}$$
for all $x$.
Now look at Equations $(1)$ and $(2)$ and add. We get $2E(x)=0$ for all $x$. It follows that $E(x)=0$ for all $x$, and therefore from Equation $(1)$, $O(x)=0$ for all $x$.
(b) Note that $\cos$ is an even function and $\sin$ is an odd function. Thus $A\sin(ax)$ is odd and $B\cos(bx)$ is even. 
It follows from (a) that if $A\sin(ax)+B\cos(bx)$ is identically $0$, then  $A\sin(ax)$ and $B\cos(bx)$ are each identically $0$. Put $x=0$. Since $\cos(0)=1$ and $\sin(0)=0$, it follows that $B=0$.
From the fact that $A\sin(ax)=0$ for all $x$, it follows that either $A=0$ or $\sin(ax)=0$ for all $x$. Suppose that $A\ne 0$. We show that $a$ must be $0$.
For if $a\ne 0$, then letting $x=\pi/2a$ we find that $\sin(\pi/2)=0$, which is false. 
A: Hint:  assume there is an $x$ such that $E(x)=a\ne0$.  What are $E(-x)$ and $O(-x)$?  How can you apply that for b)?
A: Hint $\ $ Put $\rm\:f(x) = 0\:$ in the following general decomposition into even and odd parts
$$\rm\begin{eqnarray}  &&\rm\ \ \ \ \ f(x)\, &=&\,\rm e(x) + o(x)\\
&\Rightarrow&\ \ \rm f(-x)\, &=&\,\rm e(x) - o(x)\  
\end{eqnarray}\bigg\}\   \Rightarrow \ \ 
\begin{eqnarray} \rm e(x)\, &=&\,\rm (f(x)+f(-x))/2 \\
 \rm o(x)\, &=&\,\rm (f(x)-f(-x))/2 
\end{eqnarray}$$
