For \(0^\circ \lt a^\circ \lt 90^\circ \) and \(0 \lt b \lt 1\), when \(\cos{a^\circ} = b\), which of the following expressions is equivalent to \(\cos{(2a^\circ)}\) ?

(Note: \(\cos{2\theta} = (\cos{\theta})^2-(\sin{\theta})^2\))

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Using the identity in the note:

$$ \cos{(2a^\circ)} = (\cos{a})^2 - (\sin{a})^2 $$

We need to use the pythagorean identity, which is not given:

$$ \sin^2{x}+\cos^2{x} = 1 $$ $$ \sin^2{x} = 1-\cos^2{x} $$

Substituting:

$$ (\cos{a})^2 - (\sin{a})^2 = \cos^2{a}-\sin^2{a}$$ $$ =\cos^2{a} - (1-\cos^2{a}) $$ $$ = 2\cos^2{a} - 1 $$

We are given \(\cos{a^\circ} = b\)

$$ = \boxed{2b^2 - 1} $$