Question

The maximum value of $\sin \theta + \cos \theta $ in $\left[ {0,\dfrac{\pi }{2}} \right]$ is
A.$\sqrt 2 $
B.$2$
C.$0$
D.$ - \sqrt 2 $

Answer 1

Hint: In order to find the maximum value of the given function, differentiate the function with respect to the variable given, then equate the equation with zero to get the critical points. Again, differentiate the 1st order derivative value and substitute the value of critical point in the 2nd order derivative, if it results in a negative value, then it’s the maximum or if positive then it is the minimum value.
Formula used:
$\dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }} = \cos \theta $
$\dfrac{{d\left( {\cos \theta } \right)}}{{d\theta }} = - \sin \theta $
\[\tan \dfrac{\pi }{4} = 1\]
\[\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]

Complete answer:
We are given with a function $\sin \theta + \cos \theta $ . We need to find the maximum value for the function for the interval $\left[ {0,\dfrac{\pi }{2}} \right]$.
Since, there are various methods to find the maximum value of a function, but we are solving using differentiation, which follows the steps as:
1.We would differentiate the function and then equate it with zero to find the critical points.
2.Then we would again differentiate the function obtained after first differentiation and would substitute the critical point in the function.
3.If the result obtained is less than zero, then it’s the maximum value, and if the result is greater than zero then it’s the minimum value.
Let’s follow the steps for our function given:
Considering $f\left( \theta \right) = \sin \theta + \cos \theta $
Differentiating the function with respect to $\theta $:
$f'\left( \theta \right) = \dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }} + \dfrac{{d\left( {\cos \theta } \right)}}{{d\theta }}$
Since, 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, substituting these values in the above function, we get:
$f'\left( \theta \right) = \cos \theta - \sin \theta $ ……(1)
Equating it with zero:
$f'\left( \theta \right) = \cos \theta - \sin \theta = 0$
$ \Rightarrow \cos \theta - \sin \theta = 0$
Adding both the sides by $\sin \theta $:
\[ \Rightarrow \cos \theta - \sin \theta + \sin \theta = 0 + \sin \theta \]
\[ \Rightarrow \cos \theta = \sin \theta \]
Dividing both the sides by \[\cos \theta \]:
\[ \Rightarrow \dfrac{{\cos \theta }}{{\cos \theta }} = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[ \Rightarrow 1 = \tan \theta \]
\[ \Rightarrow \tan \theta = 1\]
Since, we know that \[\tan \dfrac{\pi }{4} = 1\] , comparing this with \[\tan \theta = 1\], we get:
\[\theta = \dfrac{\pi }{4}\]
which is the critical point and also lies between $\left[ {0,\dfrac{\pi }{2}} \right]$.
Differentiating eq 1 with respect to $\theta $, we get:
$f''\left( \theta \right) = \dfrac{{d\left( {\cos \theta } \right)}}{{d\theta }} - \dfrac{{d\left( {\sin \theta } \right)}}{{d\theta }}$
$ \Rightarrow f''\left( \theta \right) = - \sin \theta - \cos \theta $
Substituting the critical point \[\theta = \dfrac{\pi }{4}\] in the above value:
\[ \Rightarrow f''\left( \theta \right) = - \sin \dfrac{\pi }{4} - \cos \dfrac{\pi }{4}\]
Substituting the values, we know that is \[\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\], we get:
\[ \Rightarrow f''\left( \theta \right) = - \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 2 }}\]

Solving it:
\[ \Rightarrow f''\left( \theta \right) = - \dfrac{2}{{\sqrt 2 }} = - \sqrt 2 \]
Since, \[f''\left( \theta \right) \leqslant 0\], so the critical point obtained is the maximum value.
Therefore, \[f\left( \theta \right)\] is maximum at \[\dfrac{\pi }{4}\].
And, the value is $f\left( {\dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4} + \cos \dfrac{\pi }{4}$.
Solving it, we get:
$ \Rightarrow f\left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow f\left( {\dfrac{\pi }{4}} \right) = \dfrac{{1 + 1}}{{\sqrt 2 }} = \dfrac{2}{{\sqrt 2 }} = \sqrt 2 $.
Therefore, the maximum value of $\sin \theta + \cos \theta $ in $\left[ {0,\dfrac{\pi }{2}} \right]$ is $\sqrt 2 $.
Hence, Option A is correct.

Note:
There are various other methods to find the maximum or minimum value for a function. Alternatively, one of them is: that for the standard function $a\sin \theta + b\cos \theta $, the maximum value obtained is equal to the square root of the sum of the square of the coefficients of the function, numerically written as $\sqrt {{a^2} + {b^2}} $.