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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 $0$.

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3 Answers 3

(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.

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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)?

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Hint $\ $ Put $\rm\:f(x) = 0\:$ in the following general decomposition into even and odd parts

$$\rm\begin{eqnarray} &&\rm\ \ \ \ \ f(x)\, &=&\,\rm e(x) + o(x)\\ &\Rightarrow&\ \ \rm f(-x)\, &=&\,\rm e(x) - o(x)\ \end{eqnarray}\bigg\}\ \Rightarrow \ \ \begin{eqnarray} \rm e(x)\, &=&\,\rm (f(x)+f(-x))/2 \\ \rm o(x)\, &=&\,\rm (f(x)-f(-x))/2 \end{eqnarray}$$

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