Evaluate cos(2pi/9) times cos(4pi/9) times cos(8pi/9).

Step-by-step solution

1. 1

   Spot the doubling pattern. Each angle is twice the one before: $\tfrac{2\pi}{9}\to\tfrac{4\pi}{9}\to\tfrac{8\pi}{9}$. Setting $x=\tfrac{2\pi}{9}$, the product is exactly $\cos x\cos2x\cos4x$ — a chain that the sine double-angle formula telescopes.

2. 2

   Recall the telescoping identity. Multiplying by $\sin x$ and applying $\sin2\theta=2\sin\theta\cos\theta$ repeatedly gives

   $\cos x\cos2x\cos4x=\frac{\sin8x}{8\sin x}$

   Each doubling contributes one factor of $2$, so three cosines produce the $8$ in the denominator.

3. 3

   Substitute $x=\tfrac{2\pi}{9}$. Then $8x=\tfrac{16\pi}{9}$ and

   $\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}=\frac{\sin\frac{16\pi}{9}}{8\sin\frac{2\pi}{9}}$

4. 4

   Simplify the sine in the numerator. Since $\tfrac{16\pi}{9}=2\pi-\tfrac{2\pi}{9}$ and $\sin(2\pi-\theta)=-\sin\theta$:

   $\sin\frac{16\pi}{9}=-\sin\frac{2\pi}{9}$

   This is the step that makes everything cancel — the numerator becomes the negative of the sine already in the denominator.

5. 5

   Cancel and finish.

   $\frac{-\sin\frac{2\pi}{9}}{8\sin\frac{2\pi}{9}}=-\frac{1}{8}$

   The cancellation is legal because $\sin\tfrac{2\pi}{9}\neq0$.

6. 6

   Verify numerically and sanity-check the sign. Direct evaluation gives $0.766044\times0.173648\times(-0.939693)=-0.125$, matching $-\tfrac18$ to within $10^{-15}$ ✓. The sign is right because $\tfrac{8\pi}{9}$ is just short of $\pi$, putting it in the second quadrant where cosine is negative, while the other two factors are positive.