Value of 36(4cos²9°-1)(4cos²27°-1)(4cos²81°-1)(4cos²243°-1)

JEE Main 2023 — 8 April, Evening Shift. Previous Year Question.

Problem

The value of $36\big(4\cos^2 9^\circ-1\big)\big(4\cos^2 27^\circ-1\big)\big(4\cos^2 81^\circ-1\big)\big(4\cos^2 243^\circ-1\big)$ is:

Key insight. A factor like $4\cos^2\theta – 1$ doesn’t obviously simplify, but rewriting $\cos^2\theta$ via $1-\sin^2\theta$ and then multiplying and dividing by $\sin\theta$ reveals the triple-angle identity $\sin 3\theta = 3\sin\theta – 4\sin^3\theta$ hiding inside it. Once every factor is rewritten this way, the whole product telescopes — each denominator cancels the numerator of the next term — leaving almost nothing behind.

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Approach

Notice that each angle in the product is exactly three times the previous one: $9^\circ \to 27^\circ \to 81^\circ \to 243^\circ$. That pattern is the signal to look for the triple-angle identity. Rewriting $4\cos^2\theta – 1$ in terms of $\sin 3\theta$ and $\sin\theta$ turns each bracket into a ratio, and multiplying four such ratios together causes consecutive numerators and denominators to cancel in sequence, since each angle is the “triple” of the one before it.

Solution

Step 1 — Rewrite $4\cos^2\theta – 1$ using the triple-angle identity

Starting from $\cos^2\theta = 1-\sin^2\theta$:

$$4\cos^2\theta – 1 = 4(1-\sin^2\theta) – 1 = 3 – 4\sin^2\theta$$

Multiplying and dividing by $\sin\theta$:

$$3 – 4\sin^2\theta = \frac{3\sin\theta – 4\sin^3\theta}{\sin\theta} = \frac{\sin 3\theta}{\sin\theta}$$

since $\sin 3\theta = 3\sin\theta – 4\sin^3\theta$ is the standard triple-angle formula.

Step 2 — Apply this to all four factors

$$36\left(\frac{\sin 27^\circ}{\sin 9^\circ}\right)\left(\frac{\sin 81^\circ}{\sin 27^\circ}\right)\left(\frac{\sin 243^\circ}{\sin 81^\circ}\right)\left(\frac{\sin 729^\circ}{\sin 243^\circ}\right)$$

(Each factor’s “output” angle — $3\theta$ — becomes the next factor’s “input” angle, since $27^\circ = 3\times 9^\circ$, $81^\circ = 3\times 27^\circ$, and so on.)

Step 3 — Cancel the telescoping terms

$\sin 27^\circ$, $\sin 81^\circ$, and $\sin 243^\circ$ each appear once in a numerator and once in a denominator, so they all cancel:

$$36\cdot\frac{\sin 729^\circ}{\sin 9^\circ}$$

Step 4 — Reduce $729^\circ$ to a standard angle

$$729^\circ = 2\times 360^\circ + 9^\circ$$

Since sine has period $360^\circ$:

$$\sin 729^\circ = \sin(2\times 360^\circ + 9^\circ) = \sin 9^\circ$$

Step 5 — Final cancellation

$$36\cdot\frac{\sin 9^\circ}{\sin 9^\circ} = 36\times 1 = 36$$

Answer

Common mistakes

• Not noticing the angles triple each time. Without spotting $9^\circ \to 27^\circ \to 81^\circ \to 243^\circ$ as successive triplings, there’s no obvious reason to reach for the triple-angle identity at all.
• Losing track of how many full rotations $729^\circ$ represents. $729^\circ \div 360^\circ = 2.025$, so it’s $2$ full rotations plus $9^\circ$ — miscounting this gives the wrong reference angle and breaks the final cancellation.

Practise next

• Evaluate $(4\cos^2 20^\circ – 1)(4\cos^2 60^\circ – 1)(4\cos^2 180^\circ – 1)$ using the same telescoping identity.