If K = sin(π/18) sin(5π/18) sin(7π/18), then the numerical value of K is ..........

Q: Why do we convert to degrees? Can't we just work in radians?

You absolutely can, but in competitive exams like JEE, speed and pattern recognition are key. Degrees make the relationship between $10^\circ, 50^\circ,$ and $70^\circ$ (or their cosine counterparts $80^\circ, 40^\circ, 20^\circ$) immediately visible, allowing you to spot the doubling pattern.

Q: Why switch to cosines? Is there a sine product formula?

There is, but it's much more complex. The cosine product formula is a classic JEE 'weapon' that simplifies these specific types of problems instantly. It's about choosing the path of least resistance to reach the solution efficiently.

Q: What if I don't remember the product formula?

That's okay! You can derive it using the $2\sin \theta \cos \theta = \sin 2\theta$ identity repeatedly. It takes a bit longer, but it's a fantastic way to build your mathematical intuition and ensure you never get stuck.

Question

If $K = \sin(\pi/18) \sin(5\pi/18) \sin(7\pi/18)$, then the numerical value of $K$ is ..........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression provided is $K = \sin(\frac{\pi}{18}) \sin(\frac{5\pi}{18}) \sin(\frac{7\pi}{18})$. At first glance, it appears to be a random collection of sine terms, but there is a hidden symmetry waiting to be discovered.

### Phase 1: The Radians vs. Degrees Realization

Radians are the language of calculus, but degrees often provide better intuition. Since $\pi 	ext{ radians} = 180^\circ$, we can perform the following conversions:

$$\frac{\pi}{18} = 10^\circ, \quad \frac{5\pi}{18} = 50^\circ, \quad \frac{7\pi}{18} = 70^\circ$$

Substituting these values, the expression simplifies to:

$$K = \sin 10^\circ \sin 50^\circ \sin 70^\circ$$

### Phase 2: The Transformation

We prefer working with cosines for this specific product. By invoking the complementary angle identity, $\sin 	heta = \cos(90^\circ - 	heta)$, we transform the terms:

$$\sin 10^\circ = \cos 80^\circ, \quad \sin 50^\circ = \cos 40^\circ, \quad \sin 70^\circ = \cos 20^\circ$$

Rearranging these in increasing order, we obtain:

$$K = \cos 20^\circ \cos 40^\circ \cos 80^\circ$$

Notice that the angles follow a geometric progression where each angle is double the previous one ($20^\circ, 40^\circ, 80^\circ$).

### Phase 3: The Power of the Formula

We utilize the standard trigonometric identity for a product of cosines:

$$\cos 	heta \cos 2	heta \cos 4	heta \dots \cos(2^{n-1}	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$$

In this problem, we set $	heta = 20^\circ$ and $n = 3$. Substituting these values into the formula yields:

$$K = \frac{\sin(2^3 \cdot 20^\circ)}{2^3 \sin 20^\circ} = \frac{\sin(160^\circ)}{8 \sin 20^\circ}$$

### Phase 4: The Final Cancellation

We now apply the supplementary angle identity, $\sin(180^\circ - 	heta) = \sin 	heta$. This allows us to simplify the numerator:

$$\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$$

Substituting this back into our expression for $K$:

$$K = \frac{\sin 20^\circ}{8 \sin 20^\circ}$$

The $\sin 20^\circ$ terms cancel out perfectly, leaving us with the final result:

**$K = \frac{1}{8}$**

Animated Solution for Mathematics - Trigonometry: If $K = sin (π /18) sin (5 π /18) sin (7 π /18)$ , then the numerical value of $K$ is ..........

Visualized Solution

Initial Expression

Convert Radians to Degrees

The Complementary Angle Identity

Transform Sine to Cosine

Rearrange the Expression

The Cosine Product Formula

Identify Parameters

Substitute into the Formula

Simplify the Expression

Supplementary Angle Identity

Final Calculation

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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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; } If K=sin(π/18)sin(5π/18)sin(7π/18), then the numerical value of K is ..........