Divide an acute angle into two parts in such a way that the sum of their tangents is minimum

Question

Assumed $0<\alpha<\pi/2$, I have divided the angle in two parts: $0<\beta<\alpha$ and $\alpha-\beta$ and I have considered $tan(\alpha-\beta)+\tan(\beta)$. By differentiating this expression is:

$\dfrac{1}{cos^2\beta}-\dfrac{1}{cos^2(\alpha-\beta)}$ and it is no negative if and only if $cos^2(\alpha-\beta)-\cos^2\beta\geq 0$.

So $cos^2\alpha\hspace{0.1cm}cos^2\beta+2\sin\alpha\cos\alpha\sin\beta\cos\alpha+sin^2\alpha\sin^2\beta-cos^2\beta=$

$cos^2\beta(cos^2\alpha-1)+2\sin\alpha\cos\alpha\sin\beta\cos\alpha+sin^2\alpha\sin^2\beta=$

$-sin^2\alpha\cos^2\beta+sin\alpha\cos\alpha\sin(2\beta)+\sin^2\alpha\sin^2\beta=$

$-sin^2\alpha(cos^2\beta-\sin^2(\beta))+sin\alpha\cos\alpha\sin(2\beta)=$

$-\sin^2\alpha\cos(2\beta)+sin\alpha\cos\alpha\sin(2\beta)\geq 0$.

Since $0<\alpha<\pi/2$ I have: $-\tan\alpha\cos(2\beta)+\sin(2\beta)\geq 0$ and since $0<\beta<\alpha<\pi/2 \rightarrow 0<2\beta<\pi\rightarrow \sin(2\beta)>0$, results $\dfrac{1}{\tan(2\beta)}\leq \dfrac{1}{\tan\alpha}$. How can I conclude that the minimum is for $\beta=\alpha/2$?

Answer 1

write for the first derivative $$\cos^2(\alpha-\beta)-\cos^2(\beta)=0$$ $$(\cos(\alpha-\beta)-\cos(\beta))(\cos(\alpha-\beta)+\cos(\beta))=0$$ and use that $$\cos(x)-\cos(y)=-2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \sin \left(\frac{x}{2}+\frac{y}{2}\right)$$ and $$\cos(x)+\cos(y)=2 \cos \left(\frac{x}{2}-\frac{y}{2}\right) \cos \left(\frac{x}{2}+\frac{y}{2}\right)$$

Answer 2

Factorise $\cos^2{(\alpha-\beta)}-\cos^2{\beta}$ as difference of two squares to get $$ (\cos{(\alpha-\beta)}+\cos{\beta})(\cos{(\alpha-\beta)}-\cos{\beta}). $$ Since $0<\beta<\alpha<\pi/2$, the first bracket is always positive, so the extremum will occur when $$ \cos{(\alpha-\beta)}-\cos{\beta} = 0. $$ Use the prosthaphaeresis formula $$ \cos{A}-\cos{B} = -2\sin{\frac{A+B}{2}}\sin{\frac{A-B}{2}} $$ to obtain $$ 0 = -2\sin{(\tfrac{1}{2}\alpha-\beta)}\sin{\tfrac{1}{2}\alpha}. $$ The second factor is not zero since $\alpha$ is not a multiple of $2\pi$. The first factor is zero when $\beta = \frac{1}{2}\alpha+k\pi$, and the only one of these that lies in $0<\beta<\alpha$ is $k=0$. Hence $\beta=\frac{1}{2}\alpha$.

To check this is a minimum, note that $-2\sin{(\frac{1}{2}\alpha-\beta)}\sin{\tfrac{1}{2}\alpha}$ is negative if $\beta<\frac{1}{2}\alpha$ and positive on the other side.

Answer 3

A symmetric approach . . .

Let $c={\large{\frac{\alpha}{2}}}$.

Let $f:[-c,c]\to \mathbb{R}$ be defined by $$f(x) = \tan(c+x) + \tan(c-x)$$ The goal is to prove that $f$ has minimum value at $x=0$.

Note that $f(-c) = f(c) = \tan(2c)$.

Also, $c={\large{\frac{\alpha}{2}}} \implies 0 < c < {\large{\frac{\pi}{4}}} \implies 0 < \tan(c) < 1$, hence $$f(-c) = f(c) = \tan(2c) = \frac{2\tan(c)}{1-\tan^2(c)} > 2\tan(c) = f(0)$$ so the minimum value of $f$ must occur for some $x \in (-c,c)$ for which $f'(x)=0$.

Taking the derivative and setting it equal to zero, we get \begin{align*} &f'(x) = 0\\[4pt] \implies\;&\sec^2(c+x) - sec^2(c-x)=0\\[4pt] \implies\;&\Bigl(1 + \tan^2(c+x)\Bigr)-\Bigl(1 + \tan^2(c-x)\Bigr)=0\\[4pt] \implies\;&\tan^2(c+x) - \tan^2(c-x)=0\\[4pt] \implies\;&\Bigl(\tan(c+x) + \tan(c-x)\Bigr)\Bigl(\tan(c+x) - \tan(c-x)\Bigr)=0\\[4pt] \implies\;&\tan(c+x) - \tan(c-x)=0\\[4pt] \implies\;&\tan(c+x) = \tan(c-x)\\[4pt] \implies\;&c+x = c-x\\[4pt] \implies\;&x = 0\\[4pt] \end{align*} so $x=0\;$is the only critical point in $(-c,c)$.

It follows that $f$ has minimum value at $x=0$, as was to be shown.