The value of $\cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot ..... \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$ is : Options: (A) $\frac{1}{256}$ (B) $\frac{1}{2}$ (C) $\frac{1}{512}$ (D) $\frac{1}{1024}$

### Analyzing the Setup The expression provided is a product of trigonometric terms: $$P = \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdot \dots \cdot \cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}}$$ At first glance, this appears to be a complex chain. However, the angles follow a clear pattern of powers of two in the denominator, specifically $2^2, 2^3, \dots, 2^{10}$. ### The Master Key The presence of $\cos \frac{\pi}{2^{10}}$ and $\sin \frac{\pi}{2^{10}}$ at the end of the product is the critical signal. We utilize the double-angle identity: $$\sin(2\theta) = 2\sin\theta\cos\theta$$ Rearranging this gives us the **Master Key** for this problem: $$\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$$ ### The Domino Effect We apply the identity to the tail end of the expression where $\theta = \frac{\pi}{2^{10}}$: $$\cos \frac{\pi}{2^{10}} \cdot \sin \frac{\pi}{2^{10}} = \frac{1}{2} \sin \left( 2 \cdot \frac{\pi}{2^{10}} \right) = \frac{1}{2} \sin \frac{\pi}{2^9}$$ This new sine term now matches the next cosine term in the sequence, $\cos \frac{\pi}{2^9}$. Applying the identity again results in: $$\left( \frac{1}{2} \sin \frac{\pi}{2^9} \right) \cdot \cos \frac{\pi}{2^9} = \frac{1}{2^2} \sin \frac{\pi}{2^8}$$ This process repeats for all 9 cosine terms. Each step reduces the power of two in the denominator of the angle and introduces an additional factor of $\frac{1}{2}$ to the coefficient. ### Final Calculation After 9 iterations, the expression collapses down to the final term: $$\frac{1}{2^8} \cdot \left( \cos \frac{\pi}{2^2} \cdot \sin \frac{\pi}{2^2} \right) = \frac{1}{2^8} \cdot \left( \frac{1}{2} \sin \left( 2 \cdot \frac{\pi}{2^2} \right) \right)$$ Simplifying the coefficient and the angle, we obtain: $$\frac{1}{2^9} \sin \frac{\pi}{2}$$ Since $\sin \frac{\pi}{2} = 1$, the expression simplifies to: $$\frac{1}{2^9} = \frac{1}{512}$$ The final answer is **$\frac{1}{512}$**.

JEE Main 2019 (10 January)

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: The value of $cos \frac{π}{2 ^{2}} \cdot cos \frac{π}{2 ^{3}} \cdot ..... \cdot cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}$ is :

Select Answer:

Progressive Hints

Hint 1

Look closely at the structure of the angles:

\frac{π}{2 ^{2}}, \frac{π}{2 ^{3}}, \dots, \frac{π}{2 ^{10}}

. Notice that the angles are halving as you move left. The key to unlocking this problem lies in the very last term,

sin \frac{π}{2 ^{10}}

, which acts as a catalyst for a chain reaction.

Hint 2

You need the most powerful tool in your trigonometric arsenal: the double-angle identity for sine. Recall that

sin (2 θ) = 2 sin θ cos θ

. If you rearrange this, you get

sin θ cos θ = \frac{1}{2} sin (2 θ)

. This is the 'master key' that will collapse the entire product.

Hint 3

Start from the rightmost end of the expression:

cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}

. Apply the identity here first. You will see that the resulting sine term has an angle of

\frac{π}{2 ^{9}}

, which perfectly matches the next cosine term in the sequence. This is the start of the domino effect.

Visualized Solution

Analyze the Trigonometric Product

Given expression: $cos \frac{π}{2 ^{2}} \cdot cos \frac{π}{2 ^{3}} \dots cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}$
Notice the pattern: The angles are halving as we move left, or doubling as we move right.
The sequence ends with a sine term having the exact same angle as the last cosine term.

The Double Angle Identity

Recall the double angle formula: $sin (2 θ) = 2 sin θ cos θ$
Rearranging gives: $sin θ cos θ = \frac{1}{2} sin (2 θ)$
This identity allows us to merge a sine and cosine pair of the same angle into a single sine term with double the angle.

Focus on the Tail End

Let's apply this to the last two terms: $cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}$
Here, our $θ$ is $\frac{π}{2 ^{10}}$ .
Substituting into the identity: $\frac{1}{2} sin (2 \cdot \frac{π}{2 ^{10}})$

The First Collapse

Simplify the angle: $2 \cdot \frac{π}{2 ^{10}} = \frac{π}{2 ^{9}}$
The last two terms collapse into: $\frac{1}{2} sin \frac{π}{2 ^{9}}$
Notice how the new angle perfectly matches the next cosine term in the sequence.

The Domino Effect

Multiply the result by the next term: $cos \frac{π}{2 ^{9}} \cdot (\frac{1}{2} sin \frac{π}{2 ^{9}})$
Apply the identity again: $\frac{1}{2} \cdot (\frac{1}{2} sin \frac{π}{2 ^{8}})$
This simplifies to: $\frac{1}{2 ^{2}} sin \frac{π}{2 ^{8}}$

Generalizing the Pattern

Each collapse halves the denominator's power of $2$ and multiplies the expression by $\frac{1}{2}$ .
Total number of cosine terms: From $2^{10}$ down to $2^{2}$ , there are $10 - 2 + 1 = 9$ terms.
This means the collapse operation will happen exactly $9$ times.

Reaching the Final Term

After $8$ collapses, we reach the very first term: $cos \frac{π}{2 ^{2}}$
The accumulated sine term will be: $\frac{1}{2 ^{8}} sin \frac{π}{2 ^{2}}$
The final multiplication is: $cos \frac{π}{2 ^{2}} \cdot (\frac{1}{2 ^{8}} sin \frac{π}{2 ^{2}})$

The Final Collapse

Apply the identity one last time: $\frac{1}{2 ^{8}} \cdot (\frac{1}{2} sin (2 \cdot \frac{π}{2 ^{2}}))$
Simplify the coefficient: $\frac{1}{2 ^{8}} \cdot \frac{1}{2} = \frac{1}{2 ^{9}}$
Simplify the angle: $2 \cdot \frac{π}{4} = \frac{π}{2}$
Result: $\frac{1}{2 ^{9}} sin \frac{π}{2}$

Evaluate the Final Result

We know the standard value: $sin \frac{π}{2} = 1$
The expression becomes: $\frac{1}{2 ^{9}} \cdot 1$
Calculate $2^{9}$ : $2^{9} = 512$
Final Answer: $\frac{1}{512}$

00:00 / 00:00

The Sigma Insight: Multiple and Sub-multiple Angles

Solution Diagram

Analyzing the Setup

The expression provided is a product of trigonometric terms:

P = cos \frac{π}{2 ^{2}} \cdot cos \frac{π}{2 ^{3}} \cdot \dots \cdot cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}

At first glance, this appears to be a complex chain. However, the angles follow a clear pattern of powers of two in the denominator, specifically

2^{2}, 2^{3}, \dots, 2^{10}

The Master Key

The presence of

cos \frac{π}{2 ^{10}}

and

sin \frac{π}{2 ^{10}}

at the end of the product is the critical signal. We utilize the double-angle identity:

sin (2 θ) = 2 sin θ cos θ

Rearranging this gives us the Master Key for this problem:

sin θ cos θ = \frac{1}{2} sin (2 θ)

The Domino Effect

We apply the identity to the tail end of the expression where

θ = \frac{π}{2 ^{10}}

cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}} = \frac{1}{2} sin (2 \cdot \frac{π}{2 ^{10}}) = \frac{1}{2} sin \frac{π}{2 ^{9}}

This new sine term now matches the next cosine term in the sequence,

cos \frac{π}{2 ^{9}}

. Applying the identity again results in:

(\frac{1}{2} sin \frac{π}{2 ^{9}}) \cdot cos \frac{π}{2 ^{9}} = \frac{1}{2 ^{2}} sin \frac{π}{2 ^{8}}

This process repeats for all 9 cosine terms. Each step reduces the power of two in the denominator of the angle and introduces an additional factor of

\frac{1}{2}

to the coefficient.

Final Calculation

After 9 iterations, the expression collapses down to the final term:

\frac{1}{2 ^{8}} \cdot (cos \frac{π}{2 ^{2}} \cdot sin \frac{π}{2 ^{2}}) = \frac{1}{2 ^{8}} \cdot (\frac{1}{2} sin (2 \cdot \frac{π}{2 ^{2}}))

Simplifying the coefficient and the angle, we obtain:

\frac{1}{2 ^{9}} sin \frac{π}{2}

Since

sin \frac{π}{2} = 1

, the expression simplifies to:

\frac{1}{2 ^{9}} = \frac{1}{512}

The final answer is $\frac{1}{512}$ .

Common Traps & FAQs

Wait, why do we start from the right side instead of the left? It feels counter-intuitive.

That is a brilliant observation! In mathematics, we often follow the flow of the equation. The double-angle identity

sin (2 θ) = 2 sin θ cos θ

requires a sine and cosine of the *same* angle to work. The left side of our expression is a long chain of cosines, but the right side provides the sine partner we desperately need to trigger the first collapse. We follow the path of least resistance—the path that allows us to use our tools immediately.

How do I know how many times this 'domino effect' will happen? Do I have to calculate every single step?

You don't have to calculate every step blindly! Count the terms. The powers of 2 in the denominator go from 10 down to 2. That is 9 terms in total. Each collapse consumes one cosine and one sine, producing a new sine with a doubled angle. Since there are 9 cosine terms, you will perform this operation exactly 9 times. Recognizing this pattern is the hallmark of a true JEE aspirant.

What happens if I get stuck in the middle of the calculation? Is there a risk of losing a factor of 1/2?

That is the most common trap! Every time you apply the identity

sin θ cos θ = \frac{1}{2} sin (2 θ)

, you *must* carry that

\frac{1}{2}

forward. Think of it as a tax you pay for simplifying the expression. Keep a running tally of these fractions outside the main product. If you lose track, just pause, breathe, and re-examine the last step. The pattern is consistent: each step introduces another

\frac{1}{2}

Animated Solution for Mathematics - Trigonometry: The value of $cos \frac{π}{2 ^{2}} \cdot cos \frac{π}{2 ^{3}} \cdot ..... \cdot cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}$ is :

Select Answer:

Visualized Solution

Analyze the Trigonometric Product

The Double Angle Identity

Focus on the Tail End

The First Collapse

The Domino Effect

Generalizing the Pattern

Reaching the Final Term

The Final Collapse

Evaluate the Final Result

The Sigma Insight: Multiple and Sub-multiple Angles

Analyzing the Setup

The Master Key

The Domino Effect

Final Calculation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of cos22π​⋅cos23π​⋅.....⋅cos210π​⋅sin210π​ is :

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Visualized Solution

Analyze the Trigonometric Product

The Double Angle Identity

Focus on the Tail End

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The Domino Effect

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