Question

Prove that the maximum value of the expression $\sin x+\cos x$ is $\sqrt{2}$.

Answer 1

First of all, modify the given expression by dividing and multiplying by $\sqrt{2}$ in the given expression. Now, use $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and modify the given expression as \sqrt{2}\left\\{ \cos \left( \dfrac{\pi }{4} \right)\times \sin x+\sin \left( \dfrac{\pi }{4} \right)\times \cos x \right\\} . Simplify it by using the formula, $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$ . We know the property that the maximum value of the sine function is always equal to 1. Use this property and calculate the maximum value of the expression.

Complete step-by-step solution:
According to the question, we are given a trigonometric expression and we have to find the maximum value of the given expression.
The given expression = $\sin x+\cos x$ …………………………………………(1)
For finding the maximum value of the given expression, we have to simplify it into a simpler form. We only know the maximum and minimum value of a single trigonometric function so, we need to convert the given expression into a single trigonometric function.
On dividing and multiplying by $\sqrt{2}$ in equation (1), we get
$=\sqrt{2}\times \dfrac{1}{\sqrt{2}}\left( \sin x+\cos x \right)$
$=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\times \sin x+\dfrac{1}{\sqrt{2}}\times \cos x \right)$ ………………………………………………(2)
We know that $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ ……………………………………(3)
From equation (2) and equation (3), we get
=\sqrt{2}\left\\{ \cos \left( \dfrac{\pi }{4} \right)\times \sin x+\sin \left( \dfrac{\pi }{4} \right)\times \cos x \right\\} ……………………………………………..(4)
In the above equation, we can observe that the expression must be modified into a simpler form.
We also know the formula, $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$ …………………………………………….(5)
Now, on replacing $y$ by $\dfrac{\pi }{4}$ in equation (5), we get
$\Rightarrow \sin \left( x+\dfrac{\pi }{4} \right)=\sin x\cos \dfrac{\pi }{4}+\cos x\sin \dfrac{\pi }{4}$ ………………………………………………….(6)
Now, from equation (4) and equation (6), we get
=\sqrt{2}\left\\{ \cos \left( \dfrac{\pi }{4} \right)\times \sin x+\sin \left( \dfrac{\pi }{4} \right)\times \cos x \right\\}
=\sqrt{2}\left\\{ \sin \left( x+\dfrac{\pi }{4} \right) \right\\} …………………………………………..(7)
We also know the property that the maximum value of sine function is always 1, that is $\sin x\le 1$ ……………………………………….(8)
Now, on replacing $x$ by $\left( x+\dfrac{\pi }{4} \right)$ in equation (8), we get
$\sin \left( x+\dfrac{\pi }{4} \right)\le 1$ …………………………………………………….(8)
From equation (7) and equation (8), we get
\sqrt{2}\left\\{ \sin \left( x+\dfrac{\pi }{4} \right) \right\\}\le \sqrt{2}\times 1
\sqrt{2}\left\\{ \sin \left( x+\dfrac{\pi }{4} \right) \right\\}\le \sqrt{2} …………………………………………..(9)
From the above equation, we can see that the expression $\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)$ is always less than or equal to $\sqrt{2}$ .
Therefore, the maximum value of the expression $\sin x+\cos x$ is equal to $\sqrt{2}$.

Note: Whenever this type of question appears where we are given an expression in terms of sine and cosine function that is $a\sin x+b\cos x$ . Then, the minimum and maximum value of the expression is given by the formula, $-\sqrt{{{a}^{2}}+{{b}^{2}}}$ and $\sqrt{{{a}^{2}}+{{b}^{2}}}$ respectively.