Question

Write the Maximum and Minimum values of $\cos (\cos x)$.

Answer 1

To solve this problem first we will find differentiation then we will find extreme points by putting differentiation is equal to 0. Then we will find second order derivatives at extreme points and we will check conditions for maxima and minima. Then we will find the maximum and minimum value of a given function.

Complete step-by-step answer:

Let $y=\cos (\cos x)$

Differentiation on both sides with respect to x

$\dfrac{dy}{dx}=-\sin (\cos x)\cdot -\sin x$

$\dfrac{dy}{dx}=\sin (\cos x)\cdot \sin x$

To get extreme value ,

$\dfrac{dy}{dx}=0$

$\therefore \sin (\cos x)\sin x=0$

If the product of two numbers is zero, then either of them will be equal to zero.

Case 1:

$\sin (\cos x)=0$

$\cos x=0$

$x=\dfrac{\pi }{2}$

Case 2:

$\sin x=0$

$x=0$

Now on differentiating $\dfrac{dy}{dx}=\sin (\cos x)\cdot \sin x$ using the product rule.

$\dfrac{d^2y}{dx^2}= \cos(\cos x)\cdot \sin x\cdot (-\sin x) + \sin (\cos x)\cdot \cos x$

If $x=0$ then,

$\dfrac{d^2y}{dx^2}= \cos(\cos 0)\cdot \sin 0\cdot (-\sin 0) + \sin (\cos 0)\cdot \cos 0$

$\dfrac{d^2y}{dx^2}= 0 + \sin (1)\cdot 1$

So from here we can say that $\dfrac{d^2y}{dx^2}$ is greater than 0 at x=0.

So at x=0 we will get the minimum value of $\cos (\cos x)$.

Substitute the value of ‘x’ in the function.

If $x=0$

$y=\cos (\cos x)$

$=\cos (\cos 0{}^\circ )$

$=\cos (1)$

So, The minimum value of $\cos (\cos x)$ is $\cos 1$.

Now,

If $x=\dfrac{\pi }{2}$ then

$\dfrac{d^2y}{dx^2}= \cos(\cos \dfrac{\pi }{2})\cdot \sin \dfrac{\pi }{2}\cdot (-\sin \dfrac{\pi }{2}) + \sin (\cos \dfrac{\pi }{2})\cdot \cos \dfrac{\pi }{2}$

$\dfrac{d^2y}{dx^2}= \cos(0)\cdot 1\cdot (-1) + \sin (0)\cdot 0$

$\dfrac{d^2y}{dx^2}= 1\cdot 1\cdot (-1) + 0$

So from here we can say that $\dfrac{d^2y}{dx^2}$ is less than 0 at $x=\dfrac{\pi }{2}$.

So at $x=\dfrac{\pi }{2}$ we will get the maximum value of $\cos (\cos x)$.

Substitute the value of ‘x’ in the function.

$y=\cos \left( \cos \left( \dfrac{\pi }{2} \right) \right)$

$y=\cos (0)$

$y=1$

$\therefore$ The maximum value of $\cos (\cos x)$ is $\cos{0}^{\circ}$=1.

Note: Cosine is a decreasing function from o to $\pi$. It has maximum value 1 when $x={{0}^{\circ }}$ and minimum value -1 when $x={\pi }$. So we can also say $\cos (\cos x)$ will have a maximum value when cos(x) has value 0.