<Question>
Find $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ \sin 100^\circ \sin 120^\circ \sin 140^\circ \sin 160^\circ.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the value of the product sin 20° sin 40° sin 60° sin 80° sin 100° sin 120° sin 140° sin 160°. Hmm, that's a lot of sine terms multiplied together. Let me start by recalling some trigonometric identities and properties that might help simplify this product.

First, I remember that products of sines can sometimes be simplified using product-to-sum formulas, but with eight terms, that might get really complicated. Alternatively, there might be a pattern or symmetry in the angles that can be exploited. Let me list out all the angles: 20°, 40°, 60°, 80°, 100°, 120°, 140°, 160°. I notice that these angles are all multiples of 20°, and they are spread out from 20° to 160°, which is 180° - 20°. That makes me think about complementary angles and possible identities involving supplementary angles.

<prompt_1.1> List out the given angles and identify any patterns or symmetries in the angles. </prompt_1.1>

The product is sin20 sin40 sin60 sin80 sin100 sin120 sin140 sin160. Let me see if I can pair terms using complementary or supplementary identities.

<prompt_1.2> Simplify the product by pairing terms using complementary or supplementary identities. </prompt_1.2>
-----

Alternatively, perhaps there is a formula for the product of multiple sine terms with angles in arithmetic progression. Wait, but these angles are not in arithmetic progression. Let me check: 20,40,60,80,100,120,140,160. Each term increases by 20°, so actually, they are in arithmetic progression with common difference 20°, starting at 20° and ending at 160°. But the number of terms is 8. Wait, 20°,40°,....,160°, which is 8 terms. So maybe there's a formula for the product of sines in arithmetic progression.

I recall that for the product sin(a) sin(a + d) sin(a + 2d)... sin(a + (n-1)d), there is a formula, but I don't remember it off the top of my head. Maybe it's too complicated. Alternatively, perhaps using complex numbers or roots of unity.

Alternatively, maybe we can use the identity for multiple angles. For example, sin(5θ) can be expressed as a product involving sinθ, sin(θ + 36°), etc.

<prompt_1.3> Explore the use of complex numbers or roots of unity to simplify the product. </prompt_1.3>
-----

Alternatively, let me check if there's a known product formula for multiple angles. For example, the product sinθ sin(60° - θ) sin(60° + θ) = (sin3θ)/4.

<prompt_1.4> Consider known product formulas for multiple angles to simplify the expression. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
For example, sin(θ) = cos(90° - θ), but I don't know if that helps here. Wait, let's check each angle:

- 20°: complementary is 70°, but 70° isn't in the product.
- 40°: complementary is 50°, not in the product.
- 60°: complementary is 30°, not in the product.
- 80°: complementary is 10°, not in the product.
- 100°: supplementary angle is 80°, which is in the product.
- 120°: supplementary angle is 60°, which is in the product.
- 140°: supplementary angle is 40°, which is in the product.
- 160°: supplementary angle is 20°, which is in the product.

So maybe using the identity sin(180° - θ) = sinθ. Let's see:

sin100° = sin(80°), sin120° = sin60°, sin140° = sin40°, sin160° = sin20°. So actually, the product can be rewritten as [sin20° sin40° sin60° sin80°]^2. Wait, let's check:

Original product: sin20 sin40 sin60 sin80 sin100 sin120 sin140 sin160

Substituting the supplementary angles:

= sin20 sin40 sin60 sin80 sin80 sin60 sin40 sin20

So grouping terms:

= (sin20)^2 (sin40)^2 (sin60)^2 (sin80)^2

So the original product is the square of (sin20 sin40 sin60 sin80). Therefore, if I can compute sin20 sin40 sin60 sin80, then squaring it will give the answer.
</execution_1.1>

<execution_1.2>
For example, sin(θ)sin(60° - θ)sin(60° + θ) = (sin3θ)/4, which is a known identity. But here, the angles don't seem to fit that exactly. Let me check if some of these angles can be expressed in terms of sums or differences.

Alternatively, I know that sin(180° - θ) = sinθ. So, for example, sin160° = sin(20°), sin140° = sin(40°), sin120° = sin(60°), sin100° = sin(80°). Oh! That's a useful symmetry. So the original product can be rewritten as:

sin20° sin40° sin60° sin80° sin80° sin60° sin40° sin20°.
</execution_1.2>

<execution_1.3>
But perhaps that's not directly applicable here. Let me think again.

Alternatively, maybe using the identity that the product of sines of angles that are in arithmetic progression can be expressed in terms of n and the angles. Wait, but I'm not sure. Let me check if there's a formula for the product of sines with angles θ, θ + α, θ + 2α, ...., θ + (n-1)α. But in our case, θ = 20°, α = 20°, n = 8. So maybe there's a formula for that.

Alternatively, maybe writing each sine term in terms of complex exponentials. For example, sinθ = (e^(iθ) - e^(-iθ))/(2i). But multiplying eight such terms would result in a very complicated expression. Maybe that's not the best approach here.

Alternatively, perhaps recognizing that some of these angles are supplementary or complementary. For example, sin160° = sin(180° - 20°) = sin20°. Similarly, sin140° = sin(180° - 40°) = sin40°, sin120° = sin(180° - 60°) = sin60°, and sin100° = sin(180° - 80°) = sin80°. Therefore, the original product can be written as [sin20° sin40° sin60° sin80°]^2.
</execution_1.3>

<execution_1.4>
Is that right? Let me verify. Let me recall the identity: sin3θ = 4sinθ sin(60° - θ) sin(60° + θ). Yes, that's correct. So, 4sinθ sin(60° - θ) sin(60° + θ) = sin3θ. So, sinθ sin(60° - θ) sin(60° + θ) = sin3θ /4.

Hmm, interesting. So if we can group terms in the product such that each group is of the form sinθ sin(60° - θ) sin(60° + θ), then we can apply this identity. Let's see if that's possible here.

Looking at the angles: 20°,40°,60°,80°,100°,120°,140°,160°. Let's note that 100° = 180° - 80°, 120° = 180° - 60°, 140° = 180° - 40°, 160° = 180° - 20°. So, the angles from 100° to 160° are supplementary angles to 80°,60°,40°,20°, respectively. And since sin(180° - x) = sinx, so sin100° = sin80°, sin120° = sin60°, sin140° = sin40°, sin160° = sin20°. Therefore, the original product can be written as [sin20° sin40° sin60° sin80°]^2.

Wait, that's a good point. Let me check:

Original product: sin20 sin40 sin60 sin80 sin100 sin120 sin140 sin160.

But sin100 = sin(180 - 80) = sin80,

sin120 = sin(180 -60) = sin60,

sin140 = sin(180 -40) = sin40,

sin160 = sin(180 -20) = sin20.

Therefore, the product becomes sin20 sin40 sin60 sin80 sin80 sin60 sin40 sin20.

Which is [sin20 sin40 sin60 sin80]^2.

So the original product is the square of sin20 sin40 sin60 sin80. Therefore, if I can compute sin20 sin40 sin60 sin80 first, then square the result, that might simplify the problem.
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
So now, the problem reduces to finding [sin20 sin40 sin60 sin80]^2. Let's focus on computing sin20 sin40 sin60 sin80 first.

<prompt_2.1> Compute the product sin20° sin40° sin60° sin80° using known values and identities. </prompt_2.1>
-----

Based on execution_1.4:
Let me compute sin20 sin40 sin60 sin80.

<prompt_2.2> Compute the product sin20° sin40° sin60° sin80° using known values and identities. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
I know that sin60° is √3/2, so that's a known value. The other terms, sin20°, sin40°, sin80°, are less straightforward. Maybe there's a way to pair these angles or use product identities.

Alternatively, I recall that sometimes products of sines can be expressed in terms of multiple angles. For example, there's an identity for the product sinθ sin(60° - θ) sin(60° + θ) which equals sin3θ / 4. Let me verify that:

Let's take θ = 20°, then sin20° sin40° sin80°. According to the identity, this should be sin60° / 4. Let's check:

sin20 sin40 sin80 = sin3*20 / 4 = sin60 / 4 = (√3/2)/4 = √3/8.

But let's compute sin20 sin40 sin80 numerically to check:

sin20 ≈ 0.3420,

sin40 ≈ 0.6428,

sin80 ≈ 0.9848.

Multiplying them: 0.3420 * 0.6428 ≈ 0.2195; 0.2195 * 0.9848 ≈ 0.2165.

√3/8 ≈ 0.2165. Yes, that's correct. So the identity holds.
</execution_2.1>

<execution_2.2>
Let me note that sin60° is √3/2. So that's a known value. Then, sin20 sin40 sin80. Let me see if I can compute sin20 sin40 sin80.

Again, perhaps using product identities. Let me recall that sinθ sin(60° - θ) sin(60° + θ) = (sin3θ)/4. If I set θ = 20°, then sin20 sin40 sin80 = sin20 sin(60° - 20°) sin(60° + 20°) = sin20 sin40 sin80 = (sin60°)/4. Wait, sin3θ where θ=20° is sin60°, so yes. Therefore, sin20 sin40 sin80 = sin60° /4.

Therefore, sin20 sin40 sin80 = (√3/2)/4 = √3/8. Then, multiplying by sin60° which is also √3/2, so sin20 sin40 sin60 sin80 = (√3/8) * (√3/2) = (3/16). Therefore, the product [sin20 sin40 sin60 sin80]^2 = (3/16)^2 = 9/256.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So, sin20 sin40 sin80 = sin60 / 4. Then, multiplying by sin60° (which is √3/2), we get:

<prompt_3.1> Calculate the squared product [sin20° sin40° sin60° sin80°]^2 and simplify the result. </prompt_3.1>
-----

Based on execution_2.2:
Therefore, the original product is 9/256. But let me verify this step by step to make sure.

<prompt_3.2> Verify the calculations for the product sin20° sin40° sin60° sin80° and confirm the final result. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
[sin20 sin40 sin60 sin80]^2 = [ (sin60 / 4) * sin60 ]^2 = [ ( (√3/2) / 4 ) * (√3/2) ]^2.

Wait, let me do this step by step. First, compute sin20 sin40 sin80:

= sin60 / 4 = (√3/2) / 4 = √3/8.

Then, multiply by sin60°:

= (√3/8) * (√3/2) = (3/16).

Therefore, [sin20 sin40 sin60 sin80]^2 = (3/16)^2 = 9/256.
</execution_3.1>

<execution_3.2>
First, sin20 sin40 sin80 = sin60 / 4. Let's confirm that:

Using the identity: sinθ sin(60° - θ) sin(60° + θ) = sin3θ /4.

Here, θ = 20°, so sin20 sin40 sin80 = sin60 /4. Correct.

Then, sin20 sin40 sin60 sin80 = (sin60 /4) * sin60 = ( (√3/2) /4 ) * (√3/2) = (√3/8)(√3/2) = (3/16). Then, squaring that gives 9/256. So the original product is 9/256.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait, is that right?

Wait, hold on. Let's check again.

<prompt_4.1> Verify the calculation of [sin20° sin40° sin60° sin80°]^2 and confirm the result. </prompt_4.1>
-----

Based on execution_3.2:
So the final answer is 9/256. But let me check once more.

Alternatively, maybe there's a miscalculation in the exponents. Let's re-express the steps:

<prompt_4.2> Re-evaluate the steps to ensure the final answer is correct. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
If sin20 sin40 sin80 = sin60 /4, then sin20 sin40 sin60 sin80 = (sin60 /4) * sin60 = (sin²60)/4. Since sin60 is √3/2, sin²60 is 3/4. Therefore, (3/4)/4 = 3/16. Then, squaring that gives (3/16)^2 = 9/256. So the original product is 9/256.
</execution_4.1>

<execution_4.2>
1. Original product: [sin20 sin40 sin60 sin80]^2.

2. Compute sin20 sin40 sin80:

Using identity: sinθ sin(60° - θ) sin(60° + θ) = sin3θ /4. Here, θ=20°, so sin20 sin40 sin80 = sin60° /4 = (√3/2)/4 = √3/8.

3. Then multiply by sin60° (which is √3/2):

√3/8 * √3/2 = (3/16).

4. Square the result: (3/16)^2 = 9/256.

Yes, that seems correct.

Alternatively, maybe there's a different approach. Let me try using complex numbers to verify.

Consider the roots of unity. The 15th roots of unity satisfy z^15 = 1. The equation can be written as z^15 -1 =0. The primitive 15th roots are e^(2πik/15) for k=1,2,....,14, but maybe that's too complicated. Alternatively, perhaps using the identity for sin products.

Wait, another approach: the product formula for sine:

sin(nθ) = 2^(n-1) product_{k=1}^{n-1} sin(θ + kπ/n).

But not sure. Alternatively, for multiple angles. Wait, perhaps using the identity for sin5θ. Let me think.

Alternatively, since all angles are multiples of 20°, which is 360°/18. Hmm. Alternatively, perhaps considering that 20°,40°,....,160° are angles in the first and second quadrants. Wait, but we already exploited the supplementary angle identity.

Alternatively, perhaps using the identity for sinθ sin(60° - θ) sin(60° + θ) = sin3θ /4. We used this for sin20 sin40 sin80 = sin60 /4. Then, sin60 is another term. Then, squaring the product.

Alternatively, maybe there is a mistake in assuming that the original product is [sin20 sin40 sin60 sin80]^2. Let me check again:

Original angles: 20,40,60,80,100,120,140,160. Then, sin100=sin80, sin120=sin60, sin140=sin40, sin160=sin20. So replacing each term:

sin100 = sin80,

sin120 = sin60,

sin140 = sin40,

sin160 = sin20.

Therefore, the product becomes sin20 sin40 sin60 sin80 sin80 sin60 sin40 sin20. So, grouping:

sin20 * sin20 * sin40 * sin40 * sin60 * sin60 * sin80 * sin80 = (sin20 sin40 sin60 sin80)^2.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
**Final Answer**
The value of the product is \boxed{\dfrac{9}{256}}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Thus, the final answer is $\boxed{\frac{9}{256}}$.
</Final_answer>
</think>