Prove Trig Identity 05Bc7C

1. **State the problem:** Prove that $$\cos^2(\theta + 45^\circ) - \frac{1}{2}(\cos 2\theta - \sin 2\theta) \equiv \sin^2 \theta.$$ 2. **Recall formulas:** - Use the cosine addition formula: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$$ - Use double angle formulas: $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta,$$ $$\sin 2\theta = 2 \sin \theta \cos \theta.$$ - Also, $$\sin^2 \theta = 1 - \cos^2 \theta.$$ 3. **Expand $$\cos^2(\theta + 45^\circ)$$:** First, find $$\cos(\theta + 45^\circ)$$: $$\cos(\theta + 45^\circ) = \cos \theta \cos 45^\circ - \sin \theta \sin 45^\circ = \frac{\sqrt{2}}{2} (\cos \theta - \sin \theta).$$ Then square it: $$\cos^2(\theta + 45^\circ) = \left( \frac{\sqrt{2}}{2} (\cos \theta - \sin \theta) \right)^2 = \frac{1}{2} (\cos \theta - \sin \theta)^2.$$ Expand the square: $$\frac{1}{2} (\cos^2 \theta - 2 \cos \theta \sin \theta + \sin^2 \theta) = \frac{1}{2} (\cos^2 \theta + \sin^2 \theta - 2 \cos \theta \sin \theta).$$ Since $$\cos^2 \theta + \sin^2 \theta = 1$$, this simplifies to: $$\cos^2(\theta + 45^\circ) = \frac{1}{2} (1 - 2 \cos \theta \sin \theta) = \frac{1}{2} - \cos \theta \sin \theta.$$ 4. **Simplify the second term:** $$- \frac{1}{2}(\cos 2\theta - \sin 2\theta) = - \frac{1}{2} \cos 2\theta + \frac{1}{2} \sin 2\theta.$$ Use double angle formulas: $$- \frac{1}{2} (\cos^2 \theta - \sin^2 \theta) + \frac{1}{2} (2 \sin \theta \cos \theta) = - \frac{1}{2} \cos^2 \theta + \frac{1}{2} \sin^2 \theta + \sin \theta \cos \theta.$$ 5. **Add the two parts:** $$\cos^2(\theta + 45^\circ) - \frac{1}{2}(\cos 2\theta - \sin 2\theta) = \left( \frac{1}{2} - \cos \theta \sin \theta \right) + \left( - \frac{1}{2} \cos^2 \theta + \frac{1}{2} \sin^2 \theta + \sin \theta \cos \theta \right).$$ Combine like terms: $$\frac{1}{2} - \cos \theta \sin \theta - \frac{1}{2} \cos^2 \theta + \frac{1}{2} \sin^2 \theta + \sin \theta \cos \theta = \frac{1}{2} - \frac{1}{2} \cos^2 \theta + \frac{1}{2} \sin^2 \theta + (- \cos \theta \sin \theta + \sin \theta \cos \theta).$$ The last two terms cancel: $$- \cos \theta \sin \theta + \sin \theta \cos \theta = 0.$$ So we have: $$\frac{1}{2} - \frac{1}{2} \cos^2 \theta + \frac{1}{2} \sin^2 \theta = \frac{1}{2} + \frac{1}{2} (\sin^2 \theta - \cos^2 \theta).$$ 6. **Use identity:** Recall $$\sin^2 \theta - \cos^2 \theta = - \cos 2\theta.$$ So: $$\frac{1}{2} + \frac{1}{2} (\sin^2 \theta - \cos^2 \theta) = \frac{1}{2} - \frac{1}{2} \cos 2\theta.$$ 7. **Rewrite right side:** Recall $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}.$$ Therefore: $$\frac{1}{2} - \frac{1}{2} \cos 2\theta = \sin^2 \theta.$$ 8. **Conclusion:** We have shown: $$\cos^2(\theta + 45^\circ) - \frac{1}{2}(\cos 2\theta - \sin 2\theta) = \sin^2 \theta,$$ which proves the identity as required.