Minimize $\cos(t)\cos(t-\alpha)$ How can I minimize $f(t)=\cos(t)\cos(t-\alpha)$? I guessed that the minimum is precisely halfway between the adjacent roots $\pi/2$ and $\pi/2+\alpha$. However, I'm not sure how to prove this. Is $f(t)$ sinusoidal via some trig identity?
 A: Use $\cos t \cos(t-\alpha) =\frac{1}{2}\big(\cos\alpha+\cos (2t-\alpha)\big)$
A: \begin{align}
\cos a\cos b +\sin a\sin b & = \cos(a-b) \\
\cos a\cos b -\sin a\sin b & = \cos(a+b) \\[8pt]
\hline
2\cos a\cos b \phantom{{}-\sin a\sin b} & = \cos(a-b)+\cos(a+b) \\[10pt]
\cos a\cos b \phantom{{}-\sin a\sin b} & = \frac{\cos(a-b)+\cos(a+b)}2 \\[8pt]
\cos t\cos(t-\alpha) \phantom{{}-\sin a\sin b} & = \frac{\cos\alpha + \cos(2t-\alpha)} 2
\end{align}
The maxima and minima of the last function are located where the maxima and minima of $\cos(2t-\alpha)$ are located, since $\cos\alpha$ is constant.
A: Hint: $f'(t_\min) = 0 $ and $f''(t_\min) > 0   $
A: One possible solution is:
The derivative of f(t) is
$$-\sin \left( t \right) \cos \left( t-\alpha \right) -\cos \left( t
 \right) \sin \left( t-\alpha \right) =0$$
and then
$$-\sin \left( 2\,t-\alpha \right) =0$$
This implies that
$$2\,t-\alpha=n\pi$$
where $n$ is an integer. Finally we obtain
$$t=(1/2)\,\alpha+(1/2)\,n\pi $$
A: Differentiate $f(t)$ to get: $$\sin(a-2t)$$
When this is zero, we will have found a critical point. 
This is zero for $t=1/2(a-\pi k)$ where k is an integer.
We now determine the nature of these critical points, so we find the 2nd derivative:
$$-2cos(a-2t)$$
We plug in the point we know is a turning point: $t$ into $f''(t)$ giving: $$-2cos(\pi k)$$ 
This is positive for odd values of k and negative for even values,
therefore a maximum when k is even and minima for k odd.
We are interested in the minimas so let $k=2n-1$ for integer n and obtain the set of values t where the function is a minima thus
$$t=1/2(a-\pi(2n-1))$$ with $n \in Z$.
