Problem source

The function $\mathrm{f}$ satisfies $\mathrm{f}(0)=1$ and 
\[
\mathrm{f}(x-y)=\mathrm{f}(x)\mathrm{f}(y)-\mathrm{f}(a-x)\mathrm{f}(a+y)
\]
for some fixed number $a$ and all $x$ and $y$. Without making any further assumptions about the nature of the function show that $\mathrm{f}(a)=0$. 
Show that, for all $t$, 
\begin{questionparts}
\item $\mathrm{f}(t)=\mathrm{f}(-t)$, 
\item $\mathrm{f}(2a)=-1$, 
\item $\mathrm{f}(2a-t)=-\mathrm{f}(t)$, 
\item $\mathrm{f}(4a+t)=\mathrm{f}(t)$. 
\end{questionparts}
Give an example of a non-constant function satisfying the conditions of the first paragraph with $a=\pi/2$. Give an example of an non-constant function satisfying the conditions of the first paragraph with $a=-2$.

Solution source

Let $P(x,y)$ be the statement that the functional equation holds, then:
\begin{align*}
P(0,0): && f(0) &= f(0)f(0)-f(a)f(a) \\
\Rightarrow && 1 &= 1 - f(a)^2 \\
\Rightarrow && f(a)^2 &= 0  \\
\Rightarrow && f(a) &= 0
\end{align*}
\begin{questionparts}
\item \begin{align*}
P(0,t): && f(-t) &= f(0)f(t) - f(a)f(a-t) \\
\Rightarrow && f(-t) &= f(t) - 0 \\
\Rightarrow && f(t) &= f(-t)
\end{align*}
\item  \begin{align*}
P(a,a): && f(0) &= f(a)f(a)-f(0)f(2a) \\
\Rightarrow && 1 &= 0 - f(2a) \\
\Rightarrow && f(2a) &= -1
\end{align*}
\item \begin{align*}
P(2a,t): && f(2a-t) &= f(2a)f(t) - f(-a)f(a+t) \\
\Rightarrow && f(2a-t) &= -f(t)-f(a)f(a+t) \\
&&&= -f(t)-0 \\
\Rightarrow && f(2a-t) &= -f(t)
\end{align*}
\item \begin{align*}
&& f(4a+t) &= f(2a-(-2a-t)) \\
&&&=-f(2a+t) \\
&&&=-f(2a-(-t)) \\
&&&=f(-t) \\
&&&=f(t)
\end{align*}
\end{questionparts}

Let $f(x) = \cos x$ then $f(\frac{\pi}{2}-x) = \sin x$ and $f(\frac{\pi}{2}+y) = -\sin y$ so the equation becomes
$\cos(x-y) = \cos x \cos y + \sin x \sin y$ which is the normal cosine addition formula.

Similarly, consider $f(x) = \cos \frac{\pi}{4} x$.