Re `[((cos2^(@)+isin2^(@))^(100)(cos5^(@)+isin5^(@))^(200))/((cos10^(@)+isin10^(@))^(126))]`

Answer

To solve the given complex number expression and find its real part, we can follow these steps:

**Step 1: Rewrite the expression using Euler's formula.**
The expression given is:
\[ \frac{(cos(2^\circ) + i \sin(2^\circ))^{100} (cos(5^\circ) + i \sin(5^\circ)^{200}}{(cos(10^\circ) + i \sin(10^\circ))^{126}} \]
Using Euler's formula \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\), we can rewrite the terms:
- \(z_1 = (cos(2^\circ) + i \sin(2^\circ))^{100} = (e^{i \cdot 2^\circ})^{100} = e^{i \cdot 200^\circ}\)
- \(z_2 = (cos(5^\circ) + i \sin(5^\circ))^{200} = (e^{i \cdot 5^\circ})^{200} = e^{i \cdot 1000^\circ}\)
- \(z_3 = (cos(10^\circ) + i \sin(10^\circ))^{126} = (e^{i \cdot 10^\circ})^{126} = e^{i \cdot 1260^\circ}\)

**Step 2: Combine the terms.**
Now we can combine these terms:
\[ z = \frac{z_1 \cdot z_2}{z_3} = \frac{e^{i \cdot 200^\circ} \cdot e^{i \cdot 1000^\circ}}{e^{i \cdot 1260^\circ}} = \frac{e^{i \cdot (200 + 1000)^\circ}}{e^{i \cdot 1260^\circ}} = \frac{e^{i \cdot 1200^\circ}}{e^{i \cdot 1260^\circ}} \]
This simplifies to:
\[ z = e^{i \cdot (1200^\circ - 1260^\circ)} = e^{-i \cdot 60^\circ} \]

**Step 3: Find the real part.**
Now we need to find the real part of \(e^{-i \cdot 60^\circ}\):
\[ e^{-i \cdot 60^\circ} = \cos(-60^\circ) + i \sin(-60^\circ) \]
Using the property of cosine and sine:
\[ \cos(-\theta) = \cos(\theta) \quad \text{and} \quad \sin(-\theta) = -\sin(\theta) \]
Thus:
\[ e^{-i \cdot 60^\circ} = \cos(60^\circ) - i \sin(60^\circ) \]
The real part is:
\[ \cos(60^\circ) = \frac{1}{2} \]

**Final Answer:**
The real part of the given complex number is \(\frac{1}{2}\).

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