Question

Maximum value of sinx-cosx is equal to
$$\begin{array}{rl}& A.\sqrt{2}\\ & B.1\\ & C.0\\ & D.\text{None of these}\end{array}$$

Answer 1

Hint: In this question, we need to find the maximum value of sinx-cosx. Suppose $f\left(x\right)=\sin x-\cos x$ for this, we will use a second derivative test. We will first find the derivative of a given function i.e. f'(x). Then we will put f'(x) = 0 and find the value of x. Then we will again find the derivative of the function i.e. f''(x). Using value of x found before in f''(x). We will check if the point gives maximum value or minimum value. If f''(x) < 0 then x gives maximum value and if f''(x) > 0 then x gives minimum value. Using the value of x which gives maximum value we will find maximum f(x). We will use following properties,
$\begin{array}{rl}& \left(i\right){\displaystyle \frac{d}{dx}}\sin x=\cos x\\ & \left(ii\right){\displaystyle \frac{d}{dx}}\cos x=-\sin x\\ & \left(iii\right)\tan x={\displaystyle \frac{\sin x}{\cos x}}\\ & \left(iv\right)\sin (\pi -{\displaystyle \frac{\pi }{4}})=\sin {\displaystyle \frac{\pi }{4}}\text{ and }\cos (\pi -{\displaystyle \frac{\pi }{4}})=-\cos {\displaystyle \frac{\pi }{4}}\end{array}$

Complete step by step answer:
Here we are given the function as sinx-cosx. Let us suppose it to be equal to f(x).
We get $f^{\prime }\left(x\right)={\displaystyle \frac{d}{dx}}(\sin x-\cos x)$.
Now let us find the derivative of a given function.
Differentiating both sides w.r.t x, we get $f^{\prime }\left(x\right)={\displaystyle \frac{d}{dx}}(\sin x-\cos x)$.
We know that ${\displaystyle \frac{d}{dx}}\sin x=\cos x\text{ and }{\displaystyle \frac{d}{dx}}\cos x=-\sin x$ so we get, $f^{\prime }\left(x\right)=\cos x-(-\sin x)\Rightarrow f^{\prime }\left(x\right)=\cos x+\sin x\cdots \cdots \cdots \left(2\right)$.
Now again let us find the derivative.
Differentiating w.r.t x we get $f^{″}\left(x\right)={\displaystyle \frac{d}{dx}}(\cos x+\sin x)$.
Again using ${\displaystyle \frac{d}{dx}}\sin x=\cos x\text{ and }{\displaystyle \frac{d}{dx}}\cos x=-\sin x$ we get, $f^{″}\left(x\right)=-\sin x+\cos x\cdots \cdots \cdots \left(3\right)$.
Now let us equate equation (2) to zero to find the value of x, we get $f^{\prime }\left(x\right)=0\Rightarrow \cos x+\sin x=0\Rightarrow \cos x=-\sin x$.
Dividing by cos x on both sides we get, $1={\displaystyle \frac{-\sin x}{\cos x}}$.
Using the property of $\tan \theta$ i.e. $\tan x={\displaystyle \frac{\sin x}{\cos x}}$ we get, $\tan x=-1$.
We know that $\tan \theta$ is negative in the second quadrant and $\tan {\displaystyle \frac{\pi }{4}}=1$. So in the second quadrant we will have $\tan (\pi -{\displaystyle \frac{\pi }{4}})=-\tan {\displaystyle \frac{\pi }{4}}=-1$.
Hence $\tan {\displaystyle \frac{3\pi }{4}}=-1$.
Therefore, $x={\displaystyle \frac{3\pi }{4}}$.
Let us check if it gives maximum value or not.
Putting $x={\displaystyle \frac{3\pi }{4}}$ in (3) we get, $f^{″}\left({\displaystyle \frac{3\pi }{4}}\right)=-\sin {\displaystyle \frac{3\pi }{4}}+\cos {\displaystyle \frac{3\pi }{4}}$.
It can be written as $-\sin (\pi -{\displaystyle \frac{\pi }{4}})+\cos (\pi -{\displaystyle \frac{\pi }{4}})$.
We know $\sin (\pi -\theta )=\sin \theta \text{ and }\cos (\pi -\theta )=-\cos \theta$ so we get,
$f^{″}\left({\displaystyle \frac{3\pi }{4}}\right)=-\sin {\displaystyle \frac{\pi }{4}}-\cos {\displaystyle \frac{\pi }{4}}$.
Since $\sin {\displaystyle \frac{\pi }{4}}={\displaystyle \frac{1}{\sqrt{2}}}=\cos {\displaystyle \frac{\pi }{4}}$ so we get, $f^{″}\left({\displaystyle \frac{3\pi }{4}}\right)=-{\displaystyle \frac{1}{\sqrt{2}}}-{\displaystyle \frac{1}{\sqrt{2}}}\Rightarrow f^{″}\left({\displaystyle \frac{3\pi }{4}}\right)=-{\displaystyle \frac{2}{\sqrt{2}}}<0$.
So $x={\displaystyle \frac{3\pi }{4}}$ gives the maximum value of f(x).
Putting $x={\displaystyle \frac{3\pi }{4}}$ in f(x) we get $f\left({\displaystyle \frac{3\pi }{4}}\right)=\sin {\displaystyle \frac{3\pi }{4}}-\cos {\displaystyle \frac{3\pi }{4}}$.
As evaluated earlier $\sin {\displaystyle \frac{3\pi }{4}}={\displaystyle \frac{1}{\sqrt{2}}}\text{ and }\cos {\displaystyle \frac{3\pi }{4}}=-{\displaystyle \frac{1}{\sqrt{2}}}$ so we get,
$f\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{1}{\sqrt{2}}}-(-{\displaystyle \frac{1}{\sqrt{2}}})\Rightarrow {\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{2}{\sqrt{2}}}$.
Splitting 2 into $\sqrt{2}\times \sqrt{2}$ we get,
$f\left({\displaystyle \frac{3\pi }{4}}\right)={\displaystyle \frac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}}=\sqrt{2}$.
Hence the maximum value of sinx-cosx is $\sqrt{2}$.

So, the correct answer is “Option A”.

Note: Students should know the values of $\tan {\displaystyle \frac{\pi }{4}},\sin {\displaystyle \frac{\pi }{4}},\cos {\displaystyle \frac{\pi }{4}}$ from the trigonometric ratio table. Note that there are many more values of $\theta$ for which $\tan \theta =-1$. They can be found by formula $\tan \theta =\tan \alpha \Rightarrow \theta =n\pi \pm \alpha ,n=0,1,2,\dots \dots$.