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

Did you know that the expression \( (\cos \theta + \sin \theta)^2 \) can be simplified using basic trigonometric identities? When you expand it, you get \( \cos^2 \theta + 2\sin \theta \cos \theta + \sin^2 \theta \). Since \(\cos^2 \theta + \sin^2 \theta = 1\), it becomes \( 1 + \sin 2\theta \) after recognizing that \( 2\sin \theta \cos \theta = \sin 2\theta \) — creating a nifty connection between sums and products in trigonometry! Now, how do we solve \( (\cos \theta + \sin \theta)^{2} = 1 - \sin 2 \theta \)? First, remember that we found it equals \( 1 + \sin 2\theta \), so you can set \( 1 + \sin 2\theta = 1 - \sin 2\theta \). Simplifying gives \( 2\sin 2\theta = 0 \), leading to solutions where \(\sin 2\theta = 0\). This occurs when \( 2\theta = n\pi \), or \( \theta = \frac{n\pi}{2} \), for any integer \( n \). Curious how various angles could showcase this relationship? Explorations are endless!