Question

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

Answer 1

Hint: 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.