Question

Find the maximum value of $\sin x+\cos x$ (using differentiation).

Answer 1

Hint: To solve this question we will take the help differentiation. We will differentiate the function with respect to $x$ and then will equate the function with zero. We again differentiate the function to check whether the value $x$ is positive or negative. If double differentiation value is positive it means the function is minimum at that value, while if it is negative the function will be maximum at that value of $x$.

Complete step by step solution:
This question asks us to find the maximum value of $\sin x+\cos x$. Consider the function $\sin x+\cos x$ . We get $f'\left( x \right)=\dfrac{d(\sin x+\cos x)}{dx}$ . Now let us find the derivative of the given function.
$f'\left( x \right)=\dfrac{d(\sin x)}{dx}+\dfrac{d(\cos x)}{dx}$…………………………………………………………..(i)
First differentiating $\sin x$ with respect to $x$ , we get:
$\dfrac{d\sin x}{dx}=\cos x$
Now differentiating $\cos x$ with respect to $x$ , we get:
$\dfrac{d\cos x}{dx}=-\sin x$
Putting the values in the formula, we get:
$f'\left( x \right)=\cos x-\sin x$ …………………………….(ii)
Equating equation (ii) with zero so that we can get the value of $x$, we get:
$\cos x-\sin x=0$
$\Rightarrow \cos x=\sin x$
Dividing each term with $\cos x$ in both L.H.S and R.H.S we get:
$\Rightarrow \dfrac{\cos x}{\cos x}=\dfrac{\sin x}{\cos x}$
We know that the fraction $\dfrac{\sin x}{\cos x}$ is the same as $\tan x$. So putting the same in the above equation, we get:
$\Rightarrow \tan x=1$
Now we would be analysing the property of $\tan x$. In the given question we have got positive value.
$\operatorname{Tan}x$is positive in the first and third quadrant. Now on solving $x={{\tan }^{-1}}\left( 1 \right)$ we get $\dfrac{\pi }{4}$ and $\dfrac{5\pi }{4}$ as the angles.
Now we will check whether the angle $\dfrac{\pi }{4}$ and $\dfrac{5\pi }{4}$ is maximum or not. For this we will double differentiate the given function or differentiate
$f''\left( x \right)=\dfrac{d\left( f'\left( x \right) \right)}{dx}$
$\Rightarrow \dfrac{d\left( \cos x-\sin x \right)}{dx}$
$\Rightarrow f''\left( x \right)=-\sin -\cos x$ …………………………… (iii)
Putting the value $\dfrac{\pi }{4}$ in the equation (iii) we get:
$\Rightarrow -\sin \dfrac{\pi }{4}-\cos \dfrac{\pi }{4}$
Value of $\sin \dfrac{\pi }{4}$ and $\cos \dfrac{\pi }{4}$ is $\dfrac{1}{\sqrt{2}}$
$\Rightarrow -\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}$
$\Rightarrow -\dfrac{2}{\sqrt{2}}$
$\Rightarrow -\sqrt{2}<0$
Since the value turned out to be negative, so at $\dfrac{\pi }{4}$ the function is maximum.
Putting $\dfrac{5\pi }{4}$ in place of the angle$x$in equation (iii), to check whether the function is maximum or not:
$\Rightarrow -\sin \dfrac{5\pi }{4}-\cos \dfrac{5\pi }{4}$
Now, $\dfrac{5\pi }{4}$ lies in the third quadrant, both the function $\sin x$ and $\cos x$ will be negative in the third quadrant. So the value of $\sin \dfrac{5\pi }{4}$ and $\cos \dfrac{5\pi }{4}$ is $\dfrac{-1}{\sqrt{2}}$ . On putting the value :$\Rightarrow -\left( -\dfrac{1}{\sqrt{2}} \right)-\left( -\dfrac{1}{\sqrt{2}} \right)$
$\Rightarrow \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2}>0$
Since the value is positive for double differentiation of the function it means the function is minimum at angle $\dfrac{5\pi }{4}$ .
We will put $\dfrac{\pi }{4}$ in the main function $\sin x+\cos x$ to find the maximum value. On putting the value we get:
$\Rightarrow \sin \dfrac{\pi }{4}+\cos \dfrac{\pi }{4}$
$\Rightarrow \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2}$

$\therefore $ The maximum value of $\sin x+\cos x$ is $\sqrt{2}$.

Note: To solve this question we should know the values of the angles of function $\tan \dfrac{\pi }{4},$ $\sin \dfrac{\pi }{4}$ $\cos \dfrac{\pi }{4}$ from trigonometric ratio table. Note that there are many more values of $\theta $ for which $\tan \theta =1$. We should know in which quadrant does the trigonometric function give negative or positive value.