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 $\cos{a^\circ} = b$

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