Alright, let's tackle this problem step by step. The task is to simplify the product of four cosine terms: cos(2π/15), cos(4π/15), cos(8π/15), and cos(16π/15). Hmm, okay. So first, I need to recall some trigonometric identities or properties that might help simplify such a product. Let me think.

I remember that products of cosines can sometimes be simplified using product-to-sum formulas, but those formulas usually handle two terms at a time. Since there are four terms here, applying the product-to-sum formulas multiple times might get complicated. Maybe there's a smarter way to approach this, perhaps by recognizing some pattern or symmetry in the angles.

Looking at the angles: 2π/15, 4π/15, 8π/15, and 16π/15. Wait a second, 16π/15 is greater than π, so maybe we can use the identity cos(θ) = cos(2π - θ) to express it in terms of an angle less than π. Let me check:

cos(16π/15) = cos(2π - 14π/15) = cos(14π/15). But cos(14π/15) is equal to -cos(π/15) because cos(π - x) = -cos(x). So cos(14π/15) = -cos(π/15). Therefore, cos(16π/15) = -cos(π/15). So replacing that in the original product, we have:

cos(2π/15) * cos(4π/15) * cos(8π/15) * (-cos(π/15)) = -cos(π/15) * cos(2π/15) * cos(4π/15) * cos(8π/15).

So now the problem reduces to simplifying the product of these four cosines: cos(π/15), cos(2π/15), cos(4π/15), cos(8π/15), and then multiplying by -1. Maybe this is a known product?

I recall that there's a formula for the product of cosines of angles that form a geometric progression in their arguments. Let me check that. Suppose we have a product like cos(θ) * cos(2θ) * cos(4θ) * ... where each angle is double the previous. In this case, θ = π/15, so the angles are π/15, 2π/15, 4π/15, 8π/15, 16π/15, etc. But in our case, we have four terms: up to 8π/15. Wait, but 16π/15 is outside the original product, but after converting it to -cos(π/15), we have four terms.

Is there a formula for such a product? Let me recall. There's a formula for the product of cosines of angles in geometric progression where each angle is multiplied by a common ratio. For a product like cos θ * cos 2θ * cos 4θ * ... * cos(2^{n-1}θ), the product can be expressed in terms of sin(2^n θ) divided by 2^n sin θ. Let me verify that.

Yes, I think the formula is:

cos θ * cos 2θ * cos 4θ * ... * cos(2^{n-1}θ) = sin(2^n θ) / (2^n sin θ).

Let me test this with n=2: cos θ * cos 2θ = sin(4θ)/(4 sin θ). Let's check:

Left side: cos θ * cos 2θ.

Right side: sin(4θ)/(4 sin θ). Using the identity sin(4θ) = 2 sin 2θ cos 2θ, and sin 2θ = 2 sin θ cos θ. Therefore, sin(4θ) = 2*(2 sin θ cos θ)*cos 2θ = 4 sin θ cos θ cos 2θ. So sin(4θ)/(4 sin θ) = (4 sin θ cos θ cos 2θ)/(4 sin θ) = cos θ cos 2θ. So yes, it works for n=2.

Similarly, for n=3: cos θ * cos 2θ * cos 4θ = sin(8θ)/(8 sin θ). Let's check with θ=π/15:

Left side: cos(π/15) * cos(2π/15) * cos(4π/15).

Right side: sin(8*(π/15))/ (8 sin(π/15)) = sin(8π/15)/(8 sin π/15).

But sin(8π/15) is sin(π - 7π/15) = sin(7π/15). So sin(8π/15) = sin(7π/15). Therefore, the right side is sin(7π/15)/(8 sin π/15). Hmm, but how does this relate to the left side?

Wait, perhaps there's another approach. If we take the product up to cos(8π/15), then n=4 terms: cos θ * cos 2θ * cos 4θ * cos 8θ. Then according to the formula, the product would be sin(16θ)/(16 sin θ). Let's test that.

sin(16θ)/(16 sin θ) = sin(16π/15)/(16 sin π/15). But 16π/15 is equivalent to π + π/15, so sin(16π/15) = sin(π + π/15) = -sin(π/15). Therefore, sin(16π/15)/(16 sin π/15) = (-sin π/15)/(16 sin π/15) = -1/16.

But according to the formula, the product cos(π/15) * cos(2π/15) * cos(4π/15) * cos(8π/15) = sin(16π/15)/(16 sin π/15) = -1/16. Therefore, the product of these four cosines is -1/16. Therefore, the original product is -1/16 multiplied by -1 (from the earlier conversion of cos(16π/15) to -cos(π/15)). Wait, let me check:

Original product: cos(2π/15) * cos(4π/15) * cos(8π/15) * cos(16π/15) = [cos(π/15) * cos(2π/15) * cos(4π/15) * cos(8π/15)] * (-1). Therefore, since the product inside the brackets is -1/16, multiplying by (-1) gives (-1/16)*(-1) = 1/16.

Therefore, the simplified value is 1/16. Hmm, that seems straightforward. Let me verify this again step by step.

First, we converted cos(16π/15) to -cos(π/15). Then, we considered the product of the remaining four cosines: cos(π/15), cos(2π/15), cos(4π/15), cos(8π/15). Using the formula for the product of cosines in geometric progression, the product is sin(16π/15)/(16 sin π/15). Since sin(16π/15) = -sin(π/15), the product becomes (-sin π/15)/(16 sin π/15) = -1/16. Then, multiplying by the -1 from the conversion gives 1/16. So the final answer is 1/16.

But let me check if this is correct. Let me compute the numerical values to verify.

Compute each cosine:

cos(2π/15) ≈ cos(24°) ≈ 0.9135

cos(4π/15) ≈ cos(48°) ≈ 0.6691

cos(8π/15) ≈ cos(96°) ≈ -0.1045

cos(16π/15) ≈ cos(192°) ≈ -0.9781

Multiplying them all together:

0.9135 * 0.6691 ≈ 0.6118

0.6118 * (-0.1045) ≈ -0.0638

-0.0638 * (-0.9781) ≈ 0.0625

Which is approximately 1/16 (since 1/16 = 0.0625). So the numerical result matches. Therefore, the analytical result of 1/16 is correct.

Hence, the simplified value of the given product is 1/16.

**Final Answer**
\boxed{\dfrac{1}{16}}
Therefore, the simplified value of the given product is $\boxed{\frac{1}{16}}$.
</think>