Question

What is the maximum value of $\sin \theta +\cos \theta$?

Answer 1

To find the maximum value of a given expression, take the derivative of $\sin \theta +\cos \theta$ and then equate it to 0 and then find the value of θ on simplification. Then substitute the value of θ in the given expression. The value after substituting the value of θ is the maximum value.

Complete step-by-step answer:
The trigonometric expression that we need to find the maximum value is:
$\sin \theta +\cos \theta$
To find the maximum value of the above expression, we need to find the value of θ at which this expression attains maximum value.
For finding the maxima of the given expression, we are going to take the derivative of the given expression and then equate it to 0.
Let us assume,
$\sin \theta +\cos \theta =f\left( \theta \right)$
Taking derivative on both the sides with respect to θ we get,
$\cos \theta -\sin \theta =f'\left( \theta \right)$
We know that:
The derivative of $\sin \theta$ with respect to θ is $\cos \theta$ $\left( \dfrac{d\sin \theta }{d\theta }=\cos \theta \right)$.
And the derivative of $\cos \theta$ with respect to θ is $-\sin \theta$ $\left( \dfrac{d\cos \theta }{d\theta }=-\sin \theta \right)$.
Now, equating f’ (θ) to 0 we get,
$\begin{aligned} & \cos \theta -\sin \theta =0 \\\ & \Rightarrow \cos \theta =\sin \theta \\\ & \Rightarrow \tan \theta =1 \\\ \end{aligned}$
And we know that tan θ = 1 when $\theta =\dfrac{\pi }{4},\dfrac{5\pi }{4}$ in the interval of (0, 2π).
Now, to confirm whether the point that we have got is the point of maxima or minima we have to differentiate f’ (θ) again and then substitute the points in this double derivative of f (θ) and then see whether it is positive, negative or zero.
$\begin{aligned} & \cos \theta -\sin \theta =f'\left( \theta \right) \\\ & \Rightarrow f''\left( \theta \right)=-\sin \theta -\cos \theta \\\ \end{aligned}$
Now, substituting the value of $\theta =\dfrac{\pi }{4}$ in the above equation we get,
$\begin{aligned} & f''\left( \dfrac{\pi }{4} \right)=-\sin \dfrac{\pi }{4}-\cos \dfrac{\pi }{4} \\\ & \Rightarrow f''\left( \dfrac{\pi }{4} \right)=-\left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right)=-\dfrac{2}{\sqrt{2}}=-\sqrt{2} \\\ \end{aligned}$
In the above double derivative, we have put the value of $\cos \dfrac{\pi }{4},\sin \dfrac{\pi }{4}$ as $\dfrac{1}{\sqrt{2}}$ because we know that $\cos \dfrac{\pi }{4}=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
As you can see that $f''\left( \theta \right)$ at $\dfrac{\pi }{4}$ is negative so this shows that at $\theta =\dfrac{\pi }{4}$ function $f\left( \theta \right)$ is attaining a maximum.
Similarly, we are going to check $f''\left( \theta \right)$ at $\dfrac{5\pi }{4}$.
$f''\left( \dfrac{5\pi }{4} \right)=-\sin \dfrac{5\pi }{4}-\cos \dfrac{5\pi }{4}$……….Eq. (1)
We know that:
$\begin{aligned} & \sin \dfrac{5\pi }{4}=-\dfrac{1}{\sqrt{2}} \\\ & \cos \dfrac{5\pi }{4}=-\dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
Substituting these values in eq. (1) we get,
$\begin{aligned} & f''\left( \dfrac{5\pi }{4} \right)=-\left( \sin \dfrac{5\pi }{4}+\cos \dfrac{5\pi }{4} \right) \\\ & \Rightarrow f''\left( \dfrac{5\pi }{4} \right)=-\left( -\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}} \right) \\\ & \Rightarrow f''\left( \dfrac{5\pi }{4} \right)=\dfrac{2}{\sqrt{2}}=\sqrt{2} \\\ \end{aligned}$
From the above, we can see that $f''\left( \theta \right)$ at $\dfrac{5\pi }{4}$ is positive which shows that at $\theta =\dfrac{5\pi }{4}$ function $f\left( \theta \right)$ is attaining a minimum.
Hence, only at $\theta =\dfrac{\pi }{4}$ function $f\left( \theta \right)$ is attaining a maximum.
Now, substituting these values of θ in the given expression $\sin \theta +\cos \theta$ we get,
When$\theta =\dfrac{\pi }{4}$,
$\begin{aligned} & \sin \dfrac{\pi }{4}+\cos \dfrac{\pi }{4} \\\ & =\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \\\ & =\dfrac{2}{\sqrt{2}}=\sqrt{2} \\\ \end{aligned}$
From the above, we have found the maximum value of $\sin \theta +\cos \theta$ as $\sqrt{2}$ .

Note: While writing the solutions of $\tan \theta =1$, you might not consider $\dfrac{5\pi }{4}$ as a solution and still you will get the correct answer because at $x=\dfrac{5\pi }{4}$, the expression has the minimum value. This is a lucky problem that you are not getting the maximum value from other solutions but always consider all the principal solutions of a trigonometric expression which lie in the interval (0, 2π).