Cosine Double Angle 1Add2E

1. **State the problem:** Find the exact value of the expression $$2 \cos^2(-150^\circ) - 1$$. 2. **Recall the formula:** The expression resembles the double-angle identity for cosine: $$\cos(2\theta) = 2\cos^2(\theta) - 1$$ This means: $$2\cos^2(\theta) - 1 = \cos(2\theta)$$ 3. **Apply the formula:** Let $$\theta = -150^\circ$$, then $$2\cos^2(-150^\circ) - 1 = \cos(2 \times -150^\circ) = \cos(-300^\circ)$$ 4. **Use cosine even property:** Since cosine is an even function, $$\cos(-300^\circ) = \cos(300^\circ)$$ 5. **Evaluate $$\cos(300^\circ)$$:** $$300^\circ = 360^\circ - 60^\circ$$, so $$\cos(300^\circ) = \cos(360^\circ - 60^\circ) = \cos(60^\circ) = \frac{1}{2}$$ 6. **Final answer:** $$2 \cos^2(-150^\circ) - 1 = \frac{1}{2}$$