Find the minimum value of $\sec 2A+\sec 2B$, where $A + B$ is constant. $A$ and $B$ belong to $(0,π/4)$ using graph of sec x.

Question

Find the minimum value of $\sec 2A+\sec 2B$, where $A + B$ is constant. $A$ and $B$ belong to $(0,π/4)$ using graph. Obviously one can solve this question using Lagrange multipliers or find maxima-minima using differentiation but they are very lenthy. So I tried using the graph of $ y= \sec x$. Considering that tangent of$ \sec x$ is always below the graph of sec x for 0,π/4. . But that didn't work out very well. Is it even possible to use this method to solve this question? Thanks a lot. Or is there any quicker or clever method?

edit: the existing answer is vague and not helpful.

wouldn't that give us wrong solution because it is for c=1 not for any real value of c? – swarnim Apr 7 at 7:00
The solution for c seems to me complicated, but I will try it. – Dr. Sonnhard Graubner

Answer 1

Hint: Substituting $$y=1-x$$ with $$c=1$$ we get $$f(x)=\sec(2x)+\sec(2(1-x))$$ and we get the Minimum for $$x={\frac {-2\,\sin \left( 2+\pi/8 \right) -2\,\cos \left( 3/8\,\pi \right) }{-\cos \left( 2 \right) +\cos \left( \pi/4+2 \right) }} $$

Answer 2

Let $A+B=C$ and we need to minimize, for a given $C$, the function $$f(A)=\sec (2 A)+\sec (2 A-2 C)$$ The first derivative is quite tedious but, using trigonometric identities, we have $$f'(A)=-\sec ^2(2 A) \cos (C)\sec ^2(2 A-2 C)\times$$ $$ \Big[\sin (2 A-3 C)-\sin (6 A-3 C)-3 \sin (2 A-C)+\sin (2A+C)\Big] $$ which is $0$ if $A=\frac 12 C$ (and therefore $A=B=\frac 12 C$).

We also have $$f''(A)=8 \sec ^3(2 (A-C))-4 \sec (2 (A-C))+8 \sec ^3(2 A)-4 \sec (2 A)$$ and then $$f''\left(\frac{C}{2}\right)=8 \sec (C) \left(2 \sec ^2(C)-1\right)\quad > 0 \quad \forall \quad 0 \leq C \leq \frac \pi 2$$ which confirms that we are at the minimum.

So, the minimum is $$f\left(\frac{C}{2}\right)=2 \sec (C)$$