16 sin 20° sin 40° sin 80° Is Equal To

JEE Main 2022 — 26 June, Shift 2. Previous Year Question.

Problem

$16\sin20^\circ\sin40^\circ\sin80^\circ$ is equal to:

Key insight. The three angles $20^\circ$, $40^\circ$ and $80^\circ$ are exactly $\theta$, $60^\circ-\theta$ and $60^\circ+\theta$ for $\theta=20^\circ$ — a pattern with a ready-made identity that collapses the whole product to a single sine.

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Approach

Recognize that $40^\circ=60^\circ-20^\circ$ and $80^\circ=60^\circ+20^\circ$, so the three angles fit the pattern $\theta$, $60^\circ-\theta$, $60^\circ+\theta$ with $\theta=20^\circ$. This pattern has a standard identity, $\sin(60^\circ-\theta)\sin\theta\sin(60^\circ+\theta)=\dfrac{\sin3\theta}4$, which turns a product of three sines into a single sine of a tripled angle.

Solution

Step 1 — Rewrite the angles in the θ, 60°−θ, 60°+θ pattern

With $\theta=20^\circ$:

$$40^\circ = 60^\circ-20^\circ, \qquad 80^\circ = 60^\circ+20^\circ$$

So the product becomes

$$16\sin20^\circ\sin40^\circ\sin80^\circ = 16\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)$$

Step 2 — Apply the standard identity

For any angle $\theta$:

$$\sin(60^\circ-\theta)\sin\theta\sin(60^\circ+\theta) = \frac{\sin3\theta}4$$

Substituting this in:

$$16\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta) = 16\times\frac{\sin3\theta}4 = 4\sin3\theta$$

Step 3 — Substitute θ = 20°

$$4\sin(3\times20^\circ) = 4\sin60^\circ$$

Step 4 — Evaluate

$$4\sin60^\circ = 4\times\frac{\sqrt3}2 = 2\sqrt3$$

Answer

Common mistakes

• Not recognizing the θ, 60°−θ, 60°+θ pattern at all. Without spotting that $40^\circ$ and $80^\circ$ are $60^\circ$ minus and plus $20^\circ$, the problem looks like it needs a long chain of product-to-sum conversions instead of a single identity.
• Misremembering the identity’s constant. The identity is $\sin(60^\circ-\theta)\sin\theta\sin(60^\circ+\theta)=\dfrac{\sin3\theta}4$ — dropping or mis-stating the $\dfrac14$ changes the final answer by a factor of 4.
• Forgetting to triple θ before evaluating the sine. After substituting $\theta=20^\circ$, the argument becomes $3\theta=60^\circ$, not $\theta=20^\circ$ itself — evaluating $\sin20^\circ$ at this stage instead of $\sin60^\circ$ gives a wrong final value.

Practise next

• Evaluate $16\sin10^\circ\sin50^\circ\sin70^\circ$ using the same $\theta$, $60^\circ-\theta$, $60^\circ+\theta$ identity with $\theta=10^\circ$.