Question

Using the relation $2\left( 1-\cos \right)<{{x}^{2}},x=0$ or prove that $\sin \left( \tan x \right)\ge x,\forall \in \left[ 0,\dfrac{\pi }{4} \right]$ .

Answer 1

Hint: To prove $\sin \left( \tan x \right)\ge x,\forall \in \left[ 0,\dfrac{\pi }{4} \right]$ , we will consider $\sin \left( \tan x \right)=x\Rightarrow \sin \left( \tan x \right)-x=0$ . Let us assume $f(x)=\sin \left( \tan x \right)-x$ . We will find the derivative of this function which gives \[f'\left( x \right)=\dfrac{\cos \left( \tan x \right)-{{\cos }^{2}}x}{{{\cos }^{2}}x}...\left( i \right)\] . From $2\left( 1-\cos \right)<{{x}^{2}}$ , we will get \[\cos \left( \tan x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}\] by putting $x=\tan x$ . This gives \[{{f}^{'}}\left( x \right)>\dfrac{1-\dfrac{{{\tan }^{2}}x}{2}-{{\cos }^{2}}x}{{{\cos }^{2}}x}\] . We will solve this and check the range $\left[ 0,\dfrac{\pi }{4} \right]$ , which shows that f(x) is an increasing function or $f\left( x \right)\ge 0$ $\Rightarrow \sin \left( \tan x \right)-x>0$ .

Complete step-by-step solution:
We have to prove $\sin \left( \tan x \right)\ge x,\forall \in \left[ 0,\dfrac{\pi }{4} \right]$ .
Let us consider $\sin \left( \tan x \right)=x$
$\Rightarrow \sin \left( \tan x \right)-x=0$
Let us assume $f(x)=\sin \left( \tan x \right)-x$
Now, we have to differentiate the above expression with respect to x. We know that $\dfrac{d}{dx}\left( \sin x \right)=\cos x\text{ and }\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ .
\[\Rightarrow f'\left( x \right)=\cos \left( \tan x \right){{\sec }^{2}}x-1\]
We know that $\sec x=\dfrac{1}{\cos x}$ .
\[\Rightarrow f'\left( x \right)=\dfrac{\cos \left( \tan x \right)}{{{\cos }^{2}}x}-1\]
\[\Rightarrow f'\left( x \right)=\dfrac{\cos \left( \tan x \right)-{{\cos }^{2}}x}{{{\cos }^{2}}x}...\left( i \right)\]
Let us now consider $2\left( 1-\cos \right)<{{x}^{2}}$
Let us take 2 from LHS to RHS. We will get
\[\Rightarrow 1-\cos x<\dfrac{{{x}^{2}}}{2}\]
The above equation can be written as
\[-\cos x<\dfrac{{{x}^{2}}}{2}-1\]
When we move the LHS to RHS and vise-versa, we will get
\[\cos x>-\dfrac{{{x}^{2}}}{2}+1\]
Let us put $x=\tan x$ . Then the above equation becomes
\[\cos \left( \tan x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}\]
We know that when \[\cos \left( \tan x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}\] in (i), then ${{f}^{'}}\left( x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}$ . Now, let’s substitute the above condition in (i). We will get
\[{{f}^{'}}\left( x \right)>\dfrac{1-\dfrac{{{\tan }^{2}}x}{2}-{{\cos }^{2}}x}{{{\cos }^{2}}x}\]
 We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ . Hence, the above expression becomes
\[{{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x-\dfrac{{{\tan }^{2}}x}{2}}{{{\cos }^{2}}x}\]
We know that $\tan x=\dfrac{\sin x}{\cos x}$ . Hence, we can write the above expression as
\[{{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x-\dfrac{{{\sin }^{2}}x}{2{{\cos }^{2}}x}}{{{\cos }^{2}}x}\]
Now, let’s take ${{\sin }^{2}}x$ common from the numerator.
\[\begin{align}
  & {{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x\left[ 1-\dfrac{1}{2{{\cos }^{2}}x} \right]}{{{\cos }^{2}}x} \\
 & \Rightarrow {{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x\left[ 2{{\cos }^{2}}x-1 \right]}{2{{\cos }^{4}}x} \\
\end{align}\]
We know that $\cos 2x=2{{\cos }^{2}}x-1$ .
\[\Rightarrow {{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x\cos 2x}{2{{\cos }^{4}}x}\]
We are given that x ranges from $\left[ 0,\dfrac{\pi }{4} \right]$ , that is, $0\le x\le \dfrac{\pi }{4}$ . Let us multiply this with 2. We will get
$0\le 2x\le \dfrac{\pi }{2}$
We can see that $2x\ge 0$ . Hence, \[\cos 2x\ge 0\] . This means that
 \[\begin{align}
  & \dfrac{{{\sin }^{2}}x\cos 2x}{2{{\cos }^{4}}x}\ge 0 \\
 & \Rightarrow {{f}^{'}}\left( x \right)\ge 0 \\
\end{align}\]
We can see that f(x) is an increasing function $\forall \in \left[ 0,\dfrac{\pi }{4} \right]$ .
We found that f(x) is an increasing function , that is, $f\left( x \right)\ge 0$ . This means that
$\Rightarrow \sin \left( \tan x \right)-x\ge 0$
Let us take x to RHS. We will get
$\sin \left( \tan x \right)\ge x$
Hence proved.

Note: You have to know trigonometric identities and differentiation to solve this question. You may make mistakes by writing the range as $\left[ 0,\dfrac{\pi }{4} \right]$ as $0< x <\dfrac{\pi }{4}$ . You may make mistake by writing ${{f}^{'}}\left( x \right)< 1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}$ instead of ${{f}^{'}}\left( x \right)> 1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}$ . So take care of while using the formulas and trigonometric conditions.