<Question>
Evaluate
\[\sin (\arcsin 0.4 + \arcsin 0.5) \cdot \sin (\arcsin 0.5 - \arcsin 0.4).\]
</Question>

<think>
<Plan_1>
Okay, so I need to evaluate the expression \(\sin (\arcsin 0.4 + \arcsin 0.5) \cdot \sin (\arcsin 0.5 - \arcsin 0.4)\). Hmm, let's break this down step by step. First, I remember that \(\arcsin x\) gives an angle whose sine is x, right? So, \(\arcsin 0.4\) is an angle whose sine is 0.4, and similarly for \(\arcsin 0.5\). Let me denote \(\alpha = \arcsin 0.4\) and \(\beta = \arcsin 0.5\). That might make things easier to handle.

So then, the expression becomes \(\sin(\alpha + \beta) \cdot \sin(\beta - \alpha)\). I need to compute this product. Maybe I can use trigonometric identities to simplify this. Let me recall the sine addition and subtraction formulas.

<prompt_1.1> Use trigonometric identities to simplify the expression \(\sin(\alpha + \beta) \cdot \sin(\beta - \alpha)\). </prompt_1.1>
-----

Alternatively, maybe compute the angles first and then compute the sines? Let's see. Let me check if that approach would be feasible.

<prompt_1.2> Calculate the angles \(\alpha\) and \(\beta\) and verify if the previous result using identities is correct. </prompt_1.2>

Alternatively, maybe there's another identity that can simplify the product of sines. Let me recall that \(\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\). Let's apply that here.

Let me denote \(A = \alpha + \beta\) and \(B = \beta - \alpha\). Then,

\[
\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)].
\]

<prompt_1.3> Use the identity \(\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\) to verify the result. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
The sine of a sum is \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\), and the sine of a difference is \(\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha\). Then, multiplying these two expressions together might get complicated, but maybe there's a trigonometric identity that can simplify the product of sines of sums and differences.
</execution_1.1>

<execution_1.2>
If \(\alpha = \arcsin 0.4\), then \(\alpha\) is in the range \([- \pi/2, \pi/2]\), right? Since the range of arcsin is \([- \pi/2, \pi/2]\). Similarly, \(\beta = \arcsin 0.5\) is also in that range. So both angles are between -π/2 and π/2. But since 0.4 and 0.5 are positive, both angles are in the first quadrant, between 0 and π/2. So, α and β are both acute angles.

If I can find \(\cos(\alpha + \beta)\) and \(\cos(\beta - \alpha)\), then I can compute the product. Alternatively, maybe use the product-to-sum formula as I thought before. Wait, but I already used the sum and difference formulas for sine.
</execution_1.2>

<execution_1.3>
First, compute \(A - B = (\alpha + \beta) - (\beta - \alpha) = 2\alpha\). Then, \(A + B = (\alpha + \beta) + (\beta - \alpha) = 2\beta\). Therefore,

\[
\sin(\alpha + \beta)\sin(\beta - \alpha) = \frac{1}{2}[\cos(2\alpha) - \cos(2\beta)].
\]

So, this might be another way to compute the same expression.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, but maybe instead of expanding both sines, there's a product-to-sum formula for the product of two sine functions. Let me recall that:

<prompt_2.1> Use the product-to-sum formula to simplify the expression \(\sin(\alpha + \beta) \cdot \sin(\beta - \alpha)\). </prompt_2.1>
-----

Based on execution_1.3:
Let me try this approach.

First, compute \(\cos(2\alpha)\) and \(\cos(2\beta)\).

<prompt_2.2> Calculate \(\cos(2\alpha)\) and \(\cos(2\beta)\) and verify the result using the identity \(\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\). </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
\[
\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)].
\]

Yes, that's the one. So, applying this identity here, where \(A = \alpha + \beta\) and \(B = \beta - \alpha\), the product becomes:

\[
\frac{1}{2}[\cos((\alpha + \beta) - (\beta - \alpha)) - \cos((\alpha + \beta) + (\beta - \alpha))].
\]

Simplify the arguments inside the cosines:

First term: \((\alpha + \beta) - (\beta - \alpha) = \alpha + \beta - \beta + \alpha = 2\alpha\).

Second term: \((\alpha + \beta) + (\beta - \alpha) = \alpha + \beta + \beta - \alpha = 2\beta\).

So, the expression simplifies to:

\[
\frac{1}{2}[\cos(2\alpha) - \cos(2\beta)].
\]

That's simpler! So instead of computing two sines and then their product, I can compute \(\cos(2\alpha)\) and \(\cos(2\beta)\), subtract them, and multiply by 1/2.
</execution_2.1>

<execution_2.2>
Since \(\alpha = \arcsin 0.4\), we can find \(\cos \alpha\) using the identity \(\sin^2 \theta + \cos^2 \theta = 1\). So, \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - 0.16} = \sqrt{0.84}\). Let me compute that. \(\sqrt{0.84}\) is approximately 0.9165, but exact value is \(\sqrt{21}/5\) because 0.84 is 84/100 = 21/25. Wait, 21/25 is 0.84, so \(\sqrt{21/25} = \sqrt{21}/5 ≈ 4.583/5 ≈ 0.9166\). So, \(\cos \alpha = \sqrt{21}/5\). Similarly, \(\cos \beta = \sqrt{1 - 0.25} = \sqrt{0.75} = \sqrt{3}/2 ≈ 0.8660\).

Then, \(\cos(2\alpha) = 1 - 2\sin^2 \alpha = 1 - 2*(0.16) = 1 - 0.32 = 0.68\). Alternatively, \(\cos(2\alpha) = 2\cos^2 \alpha - 1 = 2*(21/25) - 1 = 42/25 - 1 = 17/25 = 0.68\). Similarly, \(\cos(2\beta) = 1 - 2\sin^2 \beta = 1 - 2*(0.25) = 1 - 0.5 = 0.5\). So,

\[
\sin(\alpha + \beta)\sin(\beta - \alpha) = \frac{1}{2}[0.68 - 0.5] = \frac{1}{2}(0.18) = 0.09.
\]

Which matches the previous result.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
That's a good step. Now, I need to find \(\cos(2\alpha)\) and \(\cos(2\beta)\). Let me compute those.

<prompt_3.1> Calculate \(\cos(2\alpha)\) and \(\cos(2\beta)\) and find the simplified expression for \(\sin(\alpha + \beta) \cdot \sin(\beta - \alpha)\). </prompt_3.1>
-----

Based on execution_2.2:
So, the answer is 9/100, which is 0.09.

<prompt_3.2> Conclude the solution by confirming the final answer using both methods. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
First, recall the double-angle formula for cosine: \(\cos(2\theta) = 1 - 2\sin^2 \theta\) or \(2\cos^2 \theta - 1\). Since I know \(\sin \alpha = 0.4\) and \(\sin \beta = 0.5\), it might be easier to use \(\cos(2\theta) = 1 - 2\sin^2 \theta\).

So, for \(\alpha\):

\[
\cos(2\alpha) = 1 - 2\sin^2 \alpha = 1 - 2*(0.4)^2 = 1 - 2*0.16 = 1 - 0.32 = 0.68.
\]

For \(\beta\):

\[
\cos(2\beta) = 1 - 2\sin^2 \beta = 1 - 2*(0.5)^2 = 1 - 2*0.25 = 1 - 0.5 = 0.5.
\]

Therefore, substituting back into the expression:

\[
\frac{1}{2}[0.68 - 0.5] = \frac{1}{2}(0.18) = 0.09.
\]

So, the value of the original expression is 0.09.
</execution_3.1>

<execution_3.2>
Wait, but let me check if this is correct. Let me verify the steps again.

First, using the product-to-sum formula, we converted the product of sines into a difference of cosines. Then, we found \(\cos(2\alpha)\) and \(\cos(2\beta)\) using double-angle identities. Since \(\alpha = \arcsin 0.4\), \(\sin \alpha = 0.4\), so \(\sin^2 \alpha = 0.16\), hence \(\cos(2\alpha) = 1 - 2*0.16 = 0.68\). Similarly, \(\sin \beta = 0.5\), so \(\sin^2 \beta = 0.25\), hence \(\cos(2\beta) = 1 - 2*0.25 = 0.5\). Therefore, the expression becomes \(\frac{1}{2}(0.68 - 0.5) = \frac{1}{2}(0.18) = 0.09\). So, 0.09 is 9/100. So, the answer is 9/100.

Alternatively, let's check with another approach. Let's compute the original expression step by step.

First, compute \(\arcsin 0.4\). Let's denote this angle as \(\alpha\). So, \(\sin \alpha = 0.4\). Similarly, \(\arcsin 0.5\) is angle \(\beta\) with \(\sin \beta = 0.5\). Then, compute \(\sin(\alpha + \beta)\) and \(\sin(\beta - \alpha)\), then multiply them.

Compute \(\sin(\alpha + \beta)\):

\(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\).

We know \(\sin \alpha = 0.4\), \(\sin \beta = 0.5\). Compute \(\cos \alpha = \sqrt{1 - 0.16} = \sqrt{0.84}\), and \(\cos \beta = \sqrt{1 - 0.25} = \sqrt{0.75}\).

So, \(\sin(\alpha + \beta) = 0.4 * \sqrt{0.75} + \sqrt{0.84} * 0.5\).

Similarly, compute \(\sin(\beta - \alpha)\):

\(\sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha = 0.5 * \sqrt{0.84} - \sqrt{0.75} * 0.4\).

Then, multiply these two results:

[0.4 * sqrt(0.75) + sqrt(0.84) * 0.5] * [0.5 * sqrt(0.84) - 0.4 * sqrt(0.75)].

Let me compute this step by step. Let me denote sqrt(0.75) as A and sqrt(0.84) as B for simplicity.

So, the first term is 0.4A + 0.5B, and the second term is 0.5B - 0.4A. Multiplying these:

(0.4A)(0.5B) + (0.4A)(-0.4A) + (0.5B)(0.5B) + (0.5B)(-0.4A)

= 0.2AB - 0.16A² + 0.25B² - 0.2AB

The 0.2AB and -0.2AB cancel out, leaving:

-0.16A² + 0.25B²

Now, substitute back A² = 0.75 and B² = 0.84:

-0.16*0.75 + 0.25*0.84

Compute each term:

-0.16*0.75 = -0.12

0.25*0.84 = 0.21

Adding them: -0.12 + 0.21 = 0.09

So, this matches the previous result.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
Therefore, the answer is \(\boxed{\dfrac{9}{100}}\).
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the final answer is $\boxed{\frac{9}{100}}$.
</Final_answer>
</think>