To find the length of [tex]\( a^2 \)[/tex] in the isosceles triangle [tex]\( ABC \)[/tex] with [tex]\( A = \frac{\pi}{4} \)[/tex] and sides [tex]\( b = c = 4 \)[/tex], we will use the Law of Cosines.
The Law of Cosines states:
[tex]\[ a^2 = b^2 + c^2 - 2bc \cos(A) \][/tex]
Given:
- [tex]\( A = \frac{\pi}{4} \)[/tex]
- [tex]\( b = 4 \)[/tex]
- [tex]\( c = 4 \)[/tex]
Let's substitute the given values into the Law of Cosines formula:
[tex]\[ a^2 = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cdot \cos\left(\frac{\pi}{4}\right) \][/tex]
First, we need to calculate [tex]\(\cos\left(\frac{\pi}{4}\right)\)[/tex]:
[tex]\[ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \][/tex]
Now substitute this value back into the formula:
[tex]\[ a^2 = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cdot \frac{\sqrt{2}}{2} \][/tex]
Simplify the expression step-by-step:
1. Calculate the squares:
[tex]\[ 4^2 = 16 \][/tex]
[tex]\[ a^2 = 16 + 16 - 2 \cdot 4 \cdot 4 \cdot \frac{\sqrt{2}}{2} \][/tex]
2. Multiply the terms:
[tex]\[ a^2 = 16 + 16 - 2 \cdot 16 \cdot \frac{\sqrt{2}}{2} \][/tex]
[tex]\[ a^2 = 16 + 16 - 16\sqrt{2} \][/tex]
3. Combine like terms:
[tex]\[ a^2 = 32 - 16\sqrt{2} \][/tex]
Thus, the length of [tex]\( a^2 \)[/tex] is [tex]\(32 - 16\sqrt{2}\)[/tex].
Hence, the correct option is:
[tex]\[ \boxed{4^2(2-\sqrt{2})} \][/tex]
