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Question: Using the relation \(2\left( 1-\cos \right)<{{x}^{2}},x=0\) or prove that \(\sin \left( \tan x \righ...

Using the relation 2(1−cos⁡)<x2,x=02\left( 1-\cos \right)<{{x}^{2}},x=0 or prove that sin⁡(tan⁡x)≥x,∀∈[0,π4]\sin \left( \tan x \right)\ge x,\forall \in \left[ 0,\dfrac{\pi }{4} \right] .

Explanation

Solution

To prove sin⁡(tan⁡x)≥x,∀∈[0,π4]\sin \left( \tan x \right)\ge x,\forall \in \left[ 0,\dfrac{\pi }{4} \right] , we will consider sin⁡(tan⁡x)=x⇒sin⁡(tan⁡x)−x=0\sin \left( \tan x \right)=x\Rightarrow \sin \left( \tan x \right)-x=0 . Let us assume f(x)=sin⁡(tan⁡x)−xf(x)=\sin \left( \tan x \right)-x . We will find the derivative of this function which gives f′(x)=cos⁡(tan⁡x)−cos⁡2xcos⁡2x...(i)f'\left( x \right)=\dfrac{\cos \left( \tan x \right)-{{\cos }^{2}}x}{{{\cos }^{2}}x}...\left( i \right) . From 2(1−cos⁡)<x22\left( 1-\cos \right)<{{x}^{2}} , we will get cos⁡(tan⁡x)>1−(tan⁡x)22\cos \left( \tan x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2} by putting x=tan⁡xx=\tan x . This gives f′(x)>1−tan⁡2x2−cos⁡2xcos⁡2x{{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 [0,π4]\left[ 0,\dfrac{\pi }{4} \right] , which shows that f(x) is an increasing function or f(x)≥0f\left( x \right)\ge 0 ⇒sin⁡(tan⁡x)−x>0\Rightarrow \sin \left( \tan x \right)-x>0 .

Complete step-by-step solution:
We have to prove sin⁡(tan⁡x)≥x,∀∈[0,π4]\sin \left( \tan x \right)\ge x,\forall \in \left[ 0,\dfrac{\pi }{4} \right] .
Let us consider sin⁡(tan⁡x)=x\sin \left( \tan x \right)=x
⇒sin⁡(tan⁡x)−x=0\Rightarrow \sin \left( \tan x \right)-x=0
Let us assume f(x)=sin⁡(tan⁡x)−xf(x)=\sin \left( \tan x \right)-x
Now, we have to differentiate the above expression with respect to x. We know that ddx(sin⁡x)=cos⁡x and ddx(tan⁡x)=sec⁡2x\dfrac{d}{dx}\left( \sin x \right)=\cos x\text{ and }\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x .
⇒f′(x)=cos⁡(tan⁡x)sec⁡2x−1\Rightarrow f'\left( x \right)=\cos \left( \tan x \right){{\sec }^{2}}x-1
We know that sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x} .
⇒f′(x)=cos⁡(tan⁡x)cos⁡2x−1\Rightarrow f'\left( x \right)=\dfrac{\cos \left( \tan x \right)}{{{\cos }^{2}}x}-1
⇒f′(x)=cos⁡(tan⁡x)−cos⁡2xcos⁡2x...(i)\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(1−cos⁡)<x22\left( 1-\cos \right)<{{x}^{2}}
Let us take 2 from LHS to RHS. We will get
⇒1−cos⁡x<x22\Rightarrow 1-\cos x<\dfrac{{{x}^{2}}}{2}
The above equation can be written as
−cos⁡x<x22−1-\cos x<\dfrac{{{x}^{2}}}{2}-1
When we move the LHS to RHS and vise-versa, we will get
cos⁡x>−x22+1\cos x>-\dfrac{{{x}^{2}}}{2}+1
Let us put x=tan⁡xx=\tan x . Then the above equation becomes
cos⁡(tan⁡x)>1−(tan⁡x)22\cos \left( \tan x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2}
We know that when cos⁡(tan⁡x)>1−(tan⁡x)22\cos \left( \tan x \right)>1-\dfrac{{{\left( \tan x \right)}^{2}}}{2} in (i), then f′(x)>1−(tan⁡x)22{{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′(x)>1−tan⁡2x2−cos⁡2xcos⁡2x{{f}^{'}}\left( x \right)>\dfrac{1-\dfrac{{{\tan }^{2}}x}{2}-{{\cos }^{2}}x}{{{\cos }^{2}}x}
We know that cos⁡2x+sin⁡2x=1{{\cos }^{2}}x+{{\sin }^{2}}x=1 . Hence, the above expression becomes
f′(x)>sin⁡2x−tan⁡2x2cos⁡2x{{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x-\dfrac{{{\tan }^{2}}x}{2}}{{{\cos }^{2}}x}
We know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} . Hence, we can write the above expression as
f′(x)>sin⁡2x−sin⁡2x2cos⁡2xcos⁡2x{{f}^{'}}\left( x \right)>\dfrac{{{\sin }^{2}}x-\dfrac{{{\sin }^{2}}x}{2{{\cos }^{2}}x}}{{{\cos }^{2}}x}
Now, let’s take sin⁡2x{{\sin }^{2}}x common from the numerator.

& {{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{aligned}$$ We know that $\cos 2x=2{{\cos }^{2}}x-1$ . $$\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{aligned} & \dfrac{{{\sin }^{2}}x\cos 2x}{2{{\cos }^{4}}x}\ge 0 \\\ & \Rightarrow {{f}^{'}}\left( x \right)\ge 0 \\\ \end{aligned}$$ 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.