If $x = 2 \sin \theta - \sin 2\theta$ and $y = 2 \cos \theta - \cos 2\theta, \theta \in [0, 2\pi]$, then $\frac{d^2y}{dx^2}$ at $\theta = \pi$ is : Options: (A) $-\frac{3}{8}$ (B) $\frac{3}{4}$ (C) $\frac{3}{2}$ (D) $-\frac{3}{4}$

### Analyzing the Setup We are given the parametric equations: $$x = 2 \sin \theta - \sin 2\theta$$ $$y = 2 \cos \theta - \cos 2\theta$$ Our objective is to determine the second derivative $\frac{d^2y}{dx^2}$ at the specific parameter value $\theta = \pi$. ### Phase 1: The First Derivative Strategy To find $\frac{dy}{dx}$, we use the parametric rule: $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$ Differentiating $x$ and $y$ with respect to $\theta$: $$\frac{dx}{d\theta} = 2 \cos \theta - 2 \cos 2\theta = 2(\cos \theta - \cos 2\theta)$$ $$\frac{dy}{d\theta} = -2 \sin \theta + 2 \sin 2\theta = 2(\sin 2\theta - \sin \theta)$$ ### Phase 2: The Trigonometric Bridge Substituting these into our derivative formula: $$\frac{dy}{dx} = \frac{2(\sin 2\theta - \sin \theta)}{2(\cos \theta - \cos 2\theta)} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$$ Applying sum-to-product identities: $$\sin 2\theta - \sin \theta = 2 \cos\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right)$$ $$\cos \theta - \cos 2\theta = 2 \sin\left(\frac{3\theta}{2}\right) \sin\left(\frac{\theta}{2}\right)$$ Substituting these identities back into the expression, the common terms cancel out: $$\frac{dy}{dx} = \frac{2 \cos(3\theta/2) \sin(\theta/2)}{2 \sin(3\theta/2) \sin(\theta/2)} = \cot\left(\frac{3\theta}{2}\right)$$ ### Phase 3: The Second Derivative Trap To find $\frac{d^2y}{dx^2}$, we must differentiate with respect to $x$ using the chain rule: $$\frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( \cot\left(\frac{3\theta}{2}\right) \right) \cdot \frac{d\theta}{dx}$$ The derivative of $\cot\left(\frac{3\theta}{2}\right)$ is $-\frac{3}{2} \csc^2\left(\frac{3\theta}{2}\right)$. Since $\frac{d\theta}{dx} = \frac{1}{dx/d\theta}$, we have: $$\frac{d^2y}{dx^2} = -\frac{3}{2} \csc^2\left(\frac{3\theta}{2}\right) \cdot \frac{1}{2(\cos \theta - \cos 2\theta)}$$ ### Phase 4: The Final Calculation Now, we evaluate the expression at $\theta = \pi$: $$\csc^2\left(\frac{3\pi}{2}\right) = (-1)^2 = 1$$ $$2(\cos \pi - \cos 2\pi) = 2(-1 - 1) = -4$$ Combining these values: $$\frac{d^2y}{dx^2} = \left( -\frac{3}{2} \cdot 1 \right) \cdot \frac{1}{-4} = \frac{3}{8}$$ The final result is **$\frac{3}{8}$**.

Wait, why do we need to multiply by $\frac{d\theta}{dx}$ at the end? Can't I just differentiate $\frac{dy}{dx}$ with respect to $\theta$?

This is the most common pitfall in JEE calculus! Remember, you are finding the derivative with respect to $x$, not $\theta$. Since $\frac{dy}{dx}$ is a function of $\theta$, the chain rule dictates that $\frac{d}{dx}(\dots) = \frac{d}{d\theta}(\dots) \cdot \frac{d\theta}{dx}$. If you skip the $\frac{d\theta}{dx}$ term, you are essentially calculating $\frac{d^2y}{dx d\theta}$, which is not what the question asks for. Always respect the chain rule!

JEE Main 2020 (9 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ, θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is :

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the parametric equations

x = 2 sin θ - sin 2 θ

and

y = 2 cos θ - cos 2 θ

as a path traced by a point. The goal is to find the second derivative

\frac{d ^{2} y}{d x ^{2}}

. Remember that in parametric form,

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})

. The trap is to differentiate with respect to

θ

and forget the chain rule component

\frac{d θ}{d x}

Hint 2

The Arsenal: Before you start differentiating, look for trigonometric simplifications. Use the sum-to-product identities:

sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})

and

cos C - cos D = - 2 sin (\frac{C + D}{2}) sin (\frac{C - D}{2})

. This will collapse the complex expression for

\frac{d y}{d x}

into a much simpler function.

Hint 3

The Execution: Start by calculating

\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ}

. Once you simplify this to

cot (\frac{3 θ}{2})

, apply the chain rule for the second derivative:

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (cot (\frac{3 θ}{2})) \cdot \frac{1}{d x / d θ}

. Substitute

θ = π

only after you have the full expression.

Visualized Solution

\text{The Parametric Setup}

Given: $x = 2 sin θ - sin 2 θ$
Given: $y = 2 cos θ - cos 2 θ$
Goal: Find $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ .

\text{First Derivative Strategy}

Parametric Rule: $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$
We must first find the individual derivatives of $x$ and $y$ with respect to $θ$ .

\text{Differentiating } x

$x = 2 sin θ - sin 2 θ$
$\frac{d x}{d θ} = 2 cos θ - 2 cos 2 θ$
Factoring out 2: $\frac{d x}{d θ} = 2 (cos θ - cos 2 θ)$

\text{Differentiating } y

$y = 2 cos θ - cos 2 θ$
$\frac{d y}{d θ} = - 2 sin θ - (- sin 2 θ \cdot 2)$
$\frac{d y}{d θ} = - 2 sin θ + 2 sin 2 θ = 2 (sin 2 θ - sin θ)$

\text{Forming } \frac{dy}{dx}

$\frac{d y}{d x} = \frac{2 ( s i n 2 θ - s i n θ )}{2 ( c o s θ - c o s 2 θ )}$
$\frac{d y}{d x} = \frac{s i n 2 θ - s i n θ}{c o s θ - c o s 2 θ}$

\text{Trigonometric Simplification}

Numerator: $sin C - sin D = 2 cos (\frac{C + D}{2}) sin (\frac{C - D}{2})$
$sin 2 θ - sin θ = 2 cos (\frac{3 θ}{2}) sin (\frac{θ}{2})$

\text{Simplifying the Denominator}

Denominator: $cos C - cos D = 2 sin (\frac{C + D}{2}) sin (\frac{D - C}{2})$
$cos θ - cos 2 θ = 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2})$

\text{The Simplified First Derivative}

$\frac{d y}{d x} = \frac{2 c o s ( \frac{3 θ}{2} ) s i n ( \frac{θ}{2} )}{2 s i n ( \frac{3 θ}{2} ) s i n ( \frac{θ}{2} )}$
$\frac{d y}{d x} = cot (\frac{3 θ}{2})$

\text{The Second Derivative Trap}

Formula: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x})$
Chain Rule: $\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (\frac{d y}{d x}) \cdot \frac{d θ}{d x}$

\text{Executing the Second Derivative}

$\frac{d}{d θ} (cot (\frac{3 θ}{2})) = - csc^{2} (\frac{3 θ}{2}) \cdot \frac{3}{2}$
$\frac{d θ}{d x} = \frac{1}{\frac{d x}{d θ}} = \frac{1}{2 ( c o s θ - c o s 2 θ )}$

\text{Assembling the Expression}

$\frac{d ^{2} y}{d x ^{2}} = (- \frac{3}{2} csc^{2} (\frac{3 θ}{2})) \cdot \frac{1}{2 ( c o s θ - c o s 2 θ )}$
$\frac{d ^{2} y}{d x ^{2}} = \frac{- 3 c s c ^{2} ( \frac{3 θ}{2} )}{4 ( c o s θ - c o s 2 θ )}$

\text{Evaluating at } \theta = \pi

Substitute $θ = π$ :
$csc^{2} (\frac{3 π}{2}) = (- 1)^{2} = 1$
$cos (π) = - 1$
$cos (2 π) = 1$

\text{Final Calculation}

Denominator: $4 (- 1 - 1) = 4 (- 2) = - 8$
Numerator: $- 3 (1) = - 3$
$\frac{d ^{2} y}{d x ^{2}} = \frac{- 3}{- 8} = \frac{3}{8}$

\text{Conclusion}

Calculated Value: $\frac{3}{8}$
Note on Options: The closest option is $- \frac{3}{8}$ (Option A). This indicates a likely typo in the original question's sign convention.
Final Answer: $\frac{3}{8}$

00:00 / 00:00

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

We are given the parametric equations:

x = 2 sin θ - sin 2 θ

y = 2 cos θ - cos 2 θ

Our objective is to determine the second derivative

\frac{d ^{2} y}{d x ^{2}}

at the specific parameter value

θ = π

Phase 1

The First Derivative Strategy

To find

\frac{d y}{d x}

, we use the parametric rule:

\frac{d y}{d x} = \frac{d y / d θ}{d x / d θ}

Differentiating

x

and

y

with respect to

θ

\frac{d x}{d θ} = 2 cos θ - 2 cos 2 θ = 2 (cos θ - cos 2 θ)

\frac{d y}{d θ} = - 2 sin θ + 2 sin 2 θ = 2 (sin 2 θ - sin θ)

Phase 2

The Trigonometric Bridge

Substituting these into our derivative formula:

\frac{d y}{d x} = \frac{2 ( sin 2 θ - sin θ )}{2 ( cos θ - cos 2 θ )} = \frac{sin 2 θ - sin θ}{cos θ - cos 2 θ}

Applying sum-to-product identities:

sin 2 θ - sin θ = 2 cos (\frac{3 θ}{2}) sin (\frac{θ}{2})

cos θ - cos 2 θ = 2 sin (\frac{3 θ}{2}) sin (\frac{θ}{2})

Substituting these identities back into the expression, the common terms cancel out:

\frac{d y}{d x} = \frac{2 cos ( 3 θ /2 ) sin ( θ /2 )}{2 sin ( 3 θ /2 ) sin ( θ /2 )} = cot (\frac{3 θ}{2})

Phase 3

The Second Derivative Trap

To find

\frac{d ^{2} y}{d x ^{2}}

, we must differentiate with respect to

x

using the chain rule:

\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d θ} (cot (\frac{3 θ}{2})) \cdot \frac{d θ}{d x}

The derivative of

cot (\frac{3 θ}{2})

- \frac{3}{2} csc^{2} (\frac{3 θ}{2})

. Since

\frac{d θ}{d x} = \frac{1}{d x / d θ}

, we have:

\frac{d ^{2} y}{d x ^{2}} = - \frac{3}{2} csc^{2} (\frac{3 θ}{2}) \cdot \frac{1}{2 ( cos θ - cos 2 θ )}

Phase 4

The Final Calculation

Now, we evaluate the expression at

θ = π

csc^{2} (\frac{3 π}{2}) = (- 1)^{2} = 1

2 (cos π - cos 2 π) = 2 (- 1 - 1) = - 4

Combining these values:

\frac{d ^{2} y}{d x ^{2}} = (- \frac{3}{2} \cdot 1) \cdot \frac{1}{- 4} = \frac{3}{8}

The final result is $\frac{3}{8}$ .

Common Traps & FAQs

Wait, why do we need to multiply by

\frac{d θ}{d x}

at the end? Can't I just differentiate

\frac{d y}{d x}

with respect to

θ

This is the most common pitfall in JEE calculus! Remember, you are finding the derivative with respect to

x

, not

θ

. Since

\frac{d y}{d x}

is a function of

θ

, the chain rule dictates that

\frac{d}{d x} (\dots) = \frac{d}{d θ} (\dots) \cdot \frac{d θ}{d x}

. If you skip the

\frac{d θ}{d x}

term, you are essentially calculating

\frac{d ^{2} y}{d x d θ}

, which is not what the question asks for. Always respect the chain rule!

Is it really necessary to use trig identities? Can't I just use the quotient rule on

\frac{d y / d θ}{d x / d θ}

Technically, yes, you could use the quotient rule, but it would be an algebraic nightmare. You would end up with a massive expression involving

sin 2 θ

and

cos 2 θ

that would be incredibly difficult to differentiate a second time. In JEE Advanced, the 'brute force' method is often a trap designed to waste your time. The trig simplification is the elegant path that reveals the true structure of the problem.

I got

\frac{3}{8}

but the options say

- \frac{3}{8}

. Did I make a mistake?

Great catch! In competitive exams, if you have followed every step with mathematical rigor and your result differs from the options by a sign, it is often a typo in the question paper. Your calculation of

\frac{3}{8}

is correct based on the provided equations. Trust your process—if your derivation is sound, the logic holds, even if the exam paper has a minor error.

Animated Solution for Mathematics - Differentiation: If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ, θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is :

Select Answer:

Visualized Solution

\text{The Parametric Setup}

\text{First Derivative Strategy}

\text{Differentiating } x

\text{Differentiating } y

\text{Forming } \frac{dy}{dx}

\text{Trigonometric Simplification}

\text{Simplifying the Denominator}

\text{The Simplified First Derivative}

\text{The Second Derivative Trap}

\text{Executing the Second Derivative}

\text{Assembling the Expression}

\text{Evaluating at } \theta = \pi

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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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; } If x=2sinθ−sin2θ and y=2cosθ−cos2θ,θ∈[0,2π], then dx2d2y​ at θ=π is :

Select Answer:

Visualized Solution

\text{The Parametric Setup}

\text{First Derivative Strategy}

\text{Differentiating } x

\text{Differentiating } y

\text{Forming } \frac{dy}{dx}

\text{Trigonometric Simplification}

\text{Simplifying the Denominator}

\text{The Simplified First Derivative}

\text{The Second Derivative Trap}

\text{Executing the Second Derivative}

\text{Assembling the Expression}

\text{Evaluating at } \theta = \pi

\text{Final Calculation}

\text{Conclusion}

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ, θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is :

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If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to :

If $y (x) = (x^{x^{x}}), x > 0$ then $\frac{d ^{2} x}{d y ^{2}} + 20$ at $x = 1$ is equal to:

If $y = y (x)$ is an implicit function of $x$ such that $lo g_{e} (x + y) = 4 x y$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $x = 0$ is equal to .

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