To tackle this problem, we start with the given expression for [tex]\(\cos A\)[/tex]:
[tex]\[ \cos A = \frac{1}{2} \left( a + \frac{1}{a} \right) \][/tex]
Our goal is to prove that:
[tex]\[ \cos 2A = \frac{1}{2} \left( a^2 + \frac{1}{a^2} \right) \][/tex]
We will use the double-angle formula for cosine:
[tex]\[ \cos 2A = 2\cos^2 A - 1 \][/tex]
First, let's express [tex]\(\cos^2 A\)[/tex] in terms of [tex]\(a\)[/tex]. Start by squaring [tex]\(\cos A\)[/tex]:
[tex]\[ \cos^2 A = \left( \frac{1}{2} \left( a + \frac{1}{a} \right) \right)^2 \][/tex]
Expanding this:
[tex]\[ \cos^2 A = \left( \frac{1}{2} \right)^2 \left( a + \frac{1}{a} \right)^2 \][/tex]
[tex]\[ \cos^2 A = \frac{1}{4} \left( a^2 + 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^2} \right) \][/tex]
[tex]\[ \cos^2 A = \frac{1}{4} \left( a^2 + 2 + \frac{1}{a^2} \right) \][/tex]
Next, substitute this into the double-angle formula:
[tex]\[ \cos 2A = 2 \cos^2 A - 1 \][/tex]
[tex]\[ \cos 2A = 2 \cdot \frac{1}{4} \left( a^2 + 2 + \frac{1}{a^2} \right) - 1 \][/tex]
[tex]\[ \cos 2A = \frac{1}{2} \left( a^2 + 2 + \frac{1}{a^2} \right) - 1 \][/tex]
Simplify the expression:
[tex]\[ \cos 2A = \frac{1}{2} \left( a^2 + 2 + \frac{1}{a^2} \right) - \frac{2}{2} \][/tex]
[tex]\[ \cos 2A = \frac{1}{2} \left( a^2 + 2 + \frac{1}{a^2} - 2 \right) \][/tex]
[tex]\[ \cos 2A = \frac{1}{2} \left( a^2 + \frac{1}{a^2} \right) \][/tex]
Thus, we have proven that:
[tex]\[ \cos 2A = \frac{1}{2} \left( a^2 + \frac{1}{a^2} \right) \][/tex]
