Question

The maximum value of $\sin \theta +\cos \theta $ is
A. 1
B. 2
C. 3
D. $\sqrt{2}$

Answer 1

Hint: We will find the derivative of the given function $\sin \theta +\cos \theta $ and equate the result of the first derivative with zero to get the critical point. In the final step we will find the second derivative of the function and find out if it is positive, negative or zero to get the maximum value of the function. If $f''\left( x \right)<0$, then it is the positive point.

Complete step-by-step answer:
In the question, we have been given $f\left( x \right)=\sin \theta +\cos \theta $ and we have to find the maximum possible value for this function. We will use the derivative method to solve this question. In the derivative method, we will find the derivative of the given function by differentiating it. In the next step we will equate the result of the first derivative with zero to get the critical point. (Critical point is the point on which the function equates to zero) In the third step we will find the second derivative of the given function and check if it is positive or negative. If the value of the second derivative is less than zero, then it is the maxima point of the function and if the value is greater than zero, then it is the minima point of the given function.
$f''\left( x \right)<0\text{ }\left( \text{maxima} \right)$
\[f''\left( x \right)>0\text{ }\left( \text{minima} \right)\]
On differentiating the given function $f\left( x \right)$, we get,
$f\left( x \right)=\sin \theta +\cos \theta $
We know that $\dfrac{d\left( \sin \theta \right)}{d\theta }=\cos \theta $
And, $\dfrac{d\left( \cos \theta \right)}{d\theta }=-\sin \theta $
So, we get $f'\left( x \right)=\cos \theta -\sin \theta $
Now, we will equate $f'\left( x \right)$ with zero to get the critical point.
$\cos \theta -\sin \theta =0$
On dividing both sides by $\cos \theta $, we get,
$\begin{align}
  & \dfrac{\cos \theta }{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }=0 \\
 & 1-\tan \theta =0 \\
 & -\tan \theta =-1 \\
 & \tan \theta =1 \\
\end{align}$
We know that the value of $\theta $ at which the value of $\tan \theta =1$ is $45{}^\circ $. So, we get $\theta =45{}^\circ $.
It means the critical point of the function is $45{}^\circ $.
Now, we will check if the given function $\sin \theta +\cos \theta $ is the maxima or minima at that point. By double differentiating the function, we get,
$\begin{align}
  & f'\left( x \right)=\cos \theta -\sin \theta \\
 & f''\left( x \right)=\dfrac{d\left( \cos \theta \right)}{d\theta }-\dfrac{d\left( \sin \theta \right)}{d\theta } \\
 & f''\left( x \right)=-\sin \theta -\cos \theta \\
\end{align}$
Taking -1 common from RHS, we get,
$f''\left( x \right)=-\left( \sin \theta +\cos \theta \right)$
Now, we will check if $f''\left( x \right)$ is greater or smaller than zero.
$\begin{align}
  & -\left( \sin \theta +\cos \theta \right)<0 \\
 & \Rightarrow f''\left( x \right)<0 \\
\end{align}$
Which means that $\theta =45{}^\circ $ is the point of maxima because $f''\left( x \right)<0$. So, $x=45$ is the point of maxima. Hence the maximum value of the function $f\left( x \right)$ at $x=45{}^\circ $ is,
$\begin{align}
  & =\sin 45{}^\circ +\cos 45{}^\circ \\
 & =\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \\
 & =\dfrac{2}{\sqrt{2}} \\
 & =\sqrt{2} \\
\end{align}$
Therefore the maximum value of $f\left( x \right)=\sin \theta +\cos \theta $ is $\sqrt{2}$ and option D is the correct answer.

Note: The value of $\cos \theta $ and $\sin \theta $ lie between -1 and 1. We can represent it graphically as shown below,

The value of $\sin \theta $ will increase from 0 to $2\pi $ and the value of $\cos \theta $ will decrease from 0 to $2\pi $.
Thus the maximum value of $\sin \theta +\cos \theta $ will lie only in the first quadrant at $45{}^\circ $.
$\begin{align}
  & f\left( x \right)=\sin \theta +\cos \theta \\
 & =\sin 45{}^\circ +\cos 45{}^\circ \\
 & =\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \\
 & =\dfrac{2}{\sqrt{2}} \\ \end{align} $