Solution :
The value of cos 45 degrees is \(1\over \sqrt{2}\).
Proof :
Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees
\(\therefore\) BC = AB
Let AB = BC = a
Then by pythagoras theorem,
\(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = \(2a^2\)
\(\implies\) AC = \(\sqrt{2}a\)
In \(\Delta\) ABC, \(\angle\) C = 45 degrees
By using trigonometric formulas,
\(cos 45^{\circ}\) = \(base\over hypotenuse\) = \(b\over h\)
\(cos 45^{\circ}\) = side adjacent to 45 degrees/hypotenuse = \(BC\over AC\) = \(a\over \sqrt{2}a\) = \(1\over \sqrt{2}\)
Hence, the value of \(cos 45^{\circ}\) = \(1\over \sqrt{2}\)
