Find the minimum value of the trigonometric equation

Question

If $x + y = 2c$, find minimum value of $ \sec x +\sec y $ if $x,y\in(0,\pi/2)$, in terms of $c$.

I was able to solve by differentiating the equation and got the answer as 2secc. But i would like to know solution with trigonometry as base or without differentiating the equation.

Answer 1

Here is a method without differentiation.

Since $x,y\in(0,\pi/2)$

$\sec x,\sec y>0$

By using $A.M\ge G.M$, we have $$\dfrac{\sec x+\sec y}{2}\ge\sqrt{\sec x\cdot\sec y}$$ $$\sec^2x+\sec^2y+2\sec x\sec y\ge4\sec x\sec y$$ $$\sec^2x+\sec^2y-2\sec x\sec y\ge0$$ $$(\sec x-\sec y)^2\ge0$$ $$\sec x-\sec y=0$$ $$x=y$$ Since, $x+y=2c$ $$x+x=2c$$ $$x=c$$ $$\sec x+\sec y=2\sec x\mbox{ which is the minimum}$$

Answer 2

The function $f(x) = \sec x$ has $f''(x) = \sec x\tan^2 x + \sec^3 x > 0$ on $(0,\frac{\pi}{2})$. Thus $f(x)$ is convex on the indicated domain and it follows that $\sec x + \sec y \ge 2\sec\left(\frac{x+y}{2}\right)= 2\sec c$ which is the minimum value.

Answer 3

$$\frac{d}{dx}(\sec x+\sec y)=\frac{\sin x}{\cos^2x}+\frac{\sin y}{\cos^2y}\frac{dy}{dx}=0,\\ \frac{dy}{dx}=-1,\\ \frac{\sin x}{\cos^2x}=\frac{\sin y}{\cos^2y}\\ \Rightarrow x=y=c.$$ Minimum value: $2\sec c.$

Answer 4

$$u=\sec x+\sec y=\dfrac{4\cos\dfrac{x+y}2\cos\dfrac{x-y}2}{\cos(x-y)+\cos(x+y)}$$

$$u=\dfrac{4\cos C\cdot t}{2t^2-1+\cos2c}$$

which is a quadratic equation in $t=\cos\dfrac{x-y}2$

As $t$ is real, the discriminant must be $\ge0$

See Proving ${1+\cos x\over2+\sin x} < \frac43$