Is $\cos^2 x \ge 1/2$ always true for $x \in [0, \pi/4]$?

Yes, it is! Let's visualize the interval. As $x$ increases from $0$ to $\pi/4$, $\cos x$ decreases from $1$ to $1/\sqrt{2}$. Consequently, $\cos^2 x$ decreases from $1$ to $1/2$. Since the minimum value of $\cos^2 x$ in this interval is exactly $1/2$, the inequality $\cos^2 x \ge 1/2$ holds true for the entire range. It is the final piece of the puzzle!

JEE Advanced 2003

LEVELJEE Advanced

Animated Solution for Mathematics - Differentiation: Using the relation $2 (1 - cos x) < x^{2}, x \neq = 0$ or otherwise, prove that $sin (tan x) \geq x, \forall x \in [0, π /4]$

Progressive Hints

Hint 1

When you see an inequality like

sin (tan x) \geq x

, the most powerful weapon in your arsenal is to define a difference function

f (x) = sin (tan x) - x

. Your goal is to prove

f (x) \geq 0

for the given interval. Since

f (0) = 0

, what does this imply about the function's behavior if you can prove it is non-decreasing?

Hint 2

Now, look at the derivative

f^{'} (x)

. After applying the chain rule, you will get

f^{'} (x) = cos (tan x) \cdot sec^{2} x - 1

. To analyze the sign of this expression, simplify it to a single fraction:

\frac{c o s ( t a n x ) - c o s ^{2} x}{c o s ^{2} x}

. The denominator is clearly positive, so focus entirely on the numerator.

Hint 3

Use the provided hint:

2 (1 - cos θ) < θ^{2}

. Substitute

θ = tan x

to get a lower bound for

cos (tan x)

. Now, you need to show that this lower bound is greater than or equal to

cos^{2} x

. This reduces to proving

sin^{2} x \geq \frac{t a n ^{2} x}{2}

, which simplifies beautifully to

cos^{2} x \geq \frac{1}{2}

Visualized Solution

The Graphical Perspective

We need to prove: $sin (tan x) \geq x$ for $x \in [0, \frac{π}{4}]$
Graphically, the curve $y = sin (tan x)$ must lie above or on the line $y = x$ .

Defining the Function $f (x)$

Define a difference function: $f (x) = sin (tan x) - x$
Our goal is to prove $f (x) \geq 0$ for $x \in [0, \frac{π}{4}]$ .
Check the boundary value: $f (0) = sin (tan 0) - 0 = 0$ .

The Monotonicity Strategy

Since $f (0) = 0$ , if $f (x)$ is an increasing function, then $f (x) \geq 0$ for all $x > 0$ .
Condition for an increasing function: $f^{'} (x) \geq 0$ .
So, we must find the derivative $f^{'} (x)$ .

Differentiating $f (x)$

Differentiate $f (x) = sin (tan x) - x$ with respect to $x$ .
Apply the Chain Rule for $sin (tan x)$ :
$f^{'} (x) = cos (tan x) \cdot \frac{d}{d x} (tan x) - 1$
$f^{'} (x) = cos (tan x) \cdot sec^{2} x - 1$

Simplifying $f^{'} (x)$

Convert $sec^{2} x$ to $\frac{1}{c o s ^{2} x}$ :
$f^{'} (x) = \frac{c o s ( t a n x )}{c o s ^{2} x} - 1$
Take a common denominator:
$f^{'} (x) = \frac{c o s ( t a n x ) - c o s ^{2} x}{c o s ^{2} x}$

Analyzing the Denominator

Look at the denominator: $cos^{2} x$ .
For $x \in [0, \frac{π}{4}]$ , $cos x > 0$ , so $cos^{2} x > 0$ .
Therefore, the sign of $f^{'} (x)$ depends entirely on the numerator.
We need to prove: $cos (tan x) - cos^{2} x \geq 0$ .

Applying the Given Hint

The problem gives a hint: $2 (1 - cos θ) < θ^{2}$ for $θ \neq = 0$ .
Rearranging this inequality:
$2 - 2 cos θ < θ^{2} ⟹ 2 cos θ > 2 - θ^{2} ⟹ cos θ > 1 - \frac{θ ^{2}}{2}$

Substituting $θ = tan x$

Substituting $θ = tan x$ into $cos θ > 1 - \frac{θ ^{2}}{2}$ :
$cos (tan x) > 1 - \frac{t a n ^{2} x}{2}$
We need to prove $cos (tan x) \geq cos^{2} x$ .
It is sufficient to prove: $1 - \frac{t a n ^{2} x}{2} \geq cos^{2} x$

Solving $1 - \frac{t a n ^{2} x}{2} \geq cos^{2} x$

Rearrange the terms:
$1 - cos^{2} x \geq \frac{t a n ^{2} x}{2}$
Use the identity $1 - cos^{2} x = sin^{2} x$ :
$sin^{2} x \geq \frac{t a n ^{2} x}{2}$
Expand $tan^{2} x = \frac{s i n ^{2} x}{c o s ^{2} x}$ :
$sin^{2} x \geq \frac{s i n ^{2} x}{2 c o s ^{2} x}$

Finalizing the Condition

We have: $sin^{2} x \geq \frac{s i n ^{2} x}{2 c o s ^{2} x}$
For $x \in (0, \frac{π}{4}]$ , $sin^{2} x > 0$ , so we can safely divide both sides by $sin^{2} x$ :
$1 \geq \frac{1}{2 c o s ^{2} x}$
Multiply by $2 cos^{2} x$ (which is positive):
$2 cos^{2} x \geq 1 ⟹ cos^{2} x \geq \frac{1}{2}$

Conclusion of the Proof

Is $cos^{2} x \geq \frac{1}{2}$ true for $x \in [0, \frac{π}{4}]$ ?
Yes! In this interval, $cos x$ decreases from $1$ to $\frac{1}{2}$ .
So, $cos^{2} x$ decreases from $1$ to $\frac{1}{2}$ . Thus, $cos^{2} x \geq \frac{1}{2}$ is always true.
This means $f^{'} (x) \geq 0$ , so $f (x)$ is increasing.
Since $f (0) = 0$ , $f (x) \geq 0 ⟹ sin (tan x) \geq x$ .

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

To prove that

sin (tan x) \geq x

for all

x \in [0, π /4]

, we employ the Difference Function strategy. We define a function

f (x)

as:

f (x) = sin (tan x) - x

If we can demonstrate that

f (x) \geq 0

for the entire interval, the inequality is proven. Note that at the boundary

x = 0

, we have

f (0) = sin (tan 0) - 0 = 0

The Derivative Approach

To determine if

f (x)

is non-decreasing, we examine its derivative

f^{'} (x)

. Applying the chain rule to

f (x) = sin (tan x) - x

f^{'} (x) = cos (tan x) \cdot sec^{2} x - 1

Using the identity

sec^{2} x = \frac{1}{c o s ^{2} x}

, we rewrite the derivative as:

f^{'} (x) = \frac{cos ( tan x )}{cos ^{2} x} - 1 = \frac{cos ( tan x ) - cos ^{2} x}{cos ^{2} x}

Since

cos^{2} x > 0

for

x \in [0, π /4]

, the sign of

f^{'} (x)

is determined solely by the numerator

N (x) = cos (tan x) - cos^{2} x

Applying the Bound

We utilize the provided inequality

2 (1 - cos θ) < θ^{2}

, which rearranges to the useful bound:

cos θ > 1 - \frac{θ ^{2}}{2}

Substituting

θ = tan x

, we obtain:

cos (tan x) > 1 - \frac{tan ^{2} x}{2}

To prove

f^{'} (x) \geq 0

, it suffices to show that

1 - \frac{t a n ^{2} x}{2} \geq cos^{2} x

Final Verification

Rearranging the inequality

1 - \frac{t a n ^{2} x}{2} \geq cos^{2} x

leads to:

1 - cos^{2} x \geq \frac{tan ^{2} x}{2}

Using the identity

1 - cos^{2} x = sin^{2} x

and

tan^{2} x = \frac{s i n ^{2} x}{c o s ^{2} x}

, we get:

sin^{2} x \geq \frac{sin ^{2} x}{2 cos ^{2} x}

For

x \in (0, π /4]

, we divide by

sin^{2} x

to obtain

1 \geq \frac{1}{2 c o s ^{2} x}

, which simplifies to:

cos^{2} x \geq \frac{1}{2}

In the interval

[0, π /4]

cos x

decreases from

1

1/ 2

, meaning

cos^{2} x

decreases from

1

1/2

. Thus, the inequality holds,

f^{'} (x) \geq 0

, and $sin (tan x) \geq x$ is proven.

Common Traps & FAQs

Why do we define

f (x) = sin (tan x) - x

instead of just manipulating the inequality directly?

Great question! In JEE Advanced, when you are asked to prove an inequality over an interval, the 'difference function' method is the gold standard. By defining

f (x) = sin (tan x) - x

, we transform a comparison problem into a monotonicity problem. We know

f (0) = 0

, so if we prove

f^{'} (x) \geq 0

, we guarantee

f (x)

never dips below the x-axis. It turns a vague inequality into a concrete calculus problem.

Why can't we just use the Taylor series expansion for

sin (tan x)

You absolutely could! Using Taylor series is a valid and powerful approach. However, the problem explicitly provides the hint

2 (1 - cos θ) < θ^{2}

. In an exam setting, when a hint is provided, it is often the intended path to ensure rigor and avoid complex error-term analysis. Using the hint is the most efficient way to prove the inequality without getting lost in higher-order terms.

cos^{2} x \geq 1/2

always true for

x \in [0, π /4]

Yes, it is! Let's visualize the interval. As

x

increases from

0

π /4

cos x

decreases from

1

1/ 2

. Consequently,

cos^{2} x

decreases from

1

1/2

. Since the minimum value of

cos^{2} x

in this interval is exactly

1/2

, the inequality

cos^{2} x \geq 1/2

holds true for the entire range. It is the final piece of the puzzle!

Animated Solution for Mathematics - Differentiation: Using the relation $2 (1 - cos x) < x^{2}, x \neq = 0$ or otherwise, prove that $sin (tan x) \geq x, \forall x \in [0, π /4]$

Visualized Solution

The Graphical Perspective

Defining the Function $f (x)$

The Monotonicity Strategy

Differentiating $f (x)$

Simplifying $f^{'} (x)$

Analyzing the Denominator

Applying the Given Hint

Substituting $θ = tan x$

Solving $1 - \frac{t a n ^{2} x}{2} \geq cos^{2} x$

Finalizing the Condition

Conclusion of the Proof

The Sigma Insight: Monotonicity

Analyzing the Setup

The Derivative Approach

Applying the Bound

Final Verification

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Comprehension Passage

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using the relation 2(1−cosx)<x2,x=0 or otherwise, prove that sin(tanx)≥x,∀x∈[0,π/4]

Visualized Solution

The Graphical Perspective

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Monotonicity Strategy

Differentiating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Analyzing the Denominator

Applying the Given Hint

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=tanx

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1−2tan2x​≥cos2x

Finalizing the Condition

Conclusion of the Proof

The Sigma Insight: Monotonicity

Analyzing the Setup

The Derivative Approach

Applying the Bound

Final Verification

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Animated Solution for Mathematics - Differentiation: Using the relation $2 (1 - cos x) < x^{2}, x \neq = 0$ or otherwise, prove that $sin (tan x) \geq x, \forall x \in [0, π /4]$

Defining the Function $f (x)$

Differentiating $f (x)$

Simplifying $f^{'} (x)$

Substituting $θ = tan x$

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Show that $2 sin x + tan x \geq 3 x$ where $0 \leq x < π /2$ .

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If $P (1) = 0$ and $d P (x) / d x > P (x)$ for all $x \geq 1$ then prove that $P (x) > 0$ for all $x > 1$ .

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