# The maximum value of the expression \sin\theta+ \cos\theta, when \theta can take any value.

## Question:

The maximum value of the expression {eq}\sin\theta+ \cos\theta {/eq}, when {eq}\theta {/eq} can take any value.

## Maximum Value of Function:

Let us consider a bounded real-valued function {eq}f(x) {/eq} defined over the real axis.

The absolute maximum of the function is found by searching first for the local extreme points.

To this end, we set the first derivative of the function to zero.

Based on the sign of the first derivative ,we identify the local maxima and the maximum value between them

is the absolute maximum of the function.

We are given the function

{eq}f(\theta) = \sin\theta+ \cos\theta {/eq} when {eq}\theta {/eq} can take any value over the real axis.

We note that the function is bounded at infinity because it is the sum of two bounded functions.

Therefore the absolute maximum of the function is found by comparing the local maxima.

Upon setting the first derivative of the function to zero we get the critical points

{eq}f'(\theta) = \cos\theta - \sin\theta = 0 \Rightarrow 1 - \tan\theta = 0 \\ \displaystyle \Rightarrow \tan\theta = 1 \\ \displaystyle \Rightarrow \theta = \frac{(2n+1)\pi}{4},\; n=0,1,... {/eq}

In order to determine the nature of the critical points, we perform a second derivative test, i.e.

and evaluate the second derivative at those points

{eq}\displaystyle f''(\theta) = -\sin\theta - \cos\theta \\ \displaystyle f''(\frac{(2n+1)\pi}{4}) = -\sqrt{2} <0 \; n \; even \\ \displaystyle f''(\frac{(2n+1)\pi}{4}) = \sqrt{2} >0 \; n \; odd \\ {/eq}

Therefore the points {eq}\displaystyle \theta = \frac{(2n+1)\pi}{4},\; n \; even {/eq} are the local maxima of the function which also coincide with absolute maxima points of the function.

At those points the function value is equal to {eq}\displaystyle f( \frac{(2n+1)\pi}{4}) = \sqrt{2}, \; n \; even. {/eq}