So only solutions where $\cos z = 0$ and $\sin z = 1/2$, but $\cos z = 0$ at $z = 90^\circ, 270^\circ$, but $\sin(270^\circ) = -1

["Understanding When $\cos z = 0$ and $\sin z = \frac{1}{2}$: A Detailed Exploration", "When studying complex trigonometric functions, one fundamental question arises: For which values of $ z $ do both $ \cos z = 0 $ and $ \sin z = \frac{1}{2} $ hold true? At first glance, the angles $ z = 90^\circ $ and $ z = 270^\circ $ come to mind, since cosine vanishes at these points on the unit circle. However, a closer inspection reveals an important mathematical nuance involving the sine values.", "### Where Is $\cos z = 0$?", "The cosine function equals zero at angles where the x-coordinate on the unit circle is zero. These occur at:", "[\nz = 90^\circ + 180^\circ k, \quad \ ext{for integer } k\n]", "In radians, this is:", "[\nz = \frac{\pi}{2} + k\pi\n]", "So the solutions for $ \cos z = 0 $ are:", "- $ z = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots $ (i.e., odd multiples of $ \frac{\pi}{2} $).", "In particular, within the period $ [0^\circ, 360^\circ) $, we have $ z = 90^\circ $ and $ z = 270^\circ $.", "But here’s the key twist: at $ z = 270^\circ $, $ \sin z = -1 $, not $ \frac{1}{2} $.", "### The Role of $\sin z$: Why $ \cos z = 0 $ Doesn’t Guarantee $ \sin z = \frac{1}{2} $", "Let’s recall the Pythagorean identity:", "[\n\sin^2 z + \cos^2 z = 1\n]", "If $ \cos z = 0 $, then:", "[\n\sin^2 z = 1 \quad \Rightarrow \quad \sin z = \pm 1\n]", "So at every solution of $ \cos z = 0 $, the sine value must be either $ +1 $ or $ -1 $, never $ \frac{1}{2} $.", "Thus, there is no angle $ z $ for which both $ \cos z = 0 $ and $ \sin z = \frac{1}{2} $ simultaneously hold.", "### Clarifying Common Misconceptions", "Some confusion may stem from mixing trigonometric values in degrees versus radians or misapplying signs:", "- At $ z = 30^\circ, 150^\circ, 210^\circ, 330^\circ $, $ \sin z = \frac{1}{2} $, but $ \cos z <br/>\neq 0 $.\n- At $ z = 90^\circ $, $ \sin z = 1 $ — not $ \frac{1}{2} $.\n- At $ z = 270^\circ $, $ \sin z = -1 $ — again, not $ \frac{1}{2} $.", "Thus, although $ \cos z = 0 $ at $ 90^\circ $ and $ 270^\circ $, the sine condition fails in both cases.", "### Solving the Equations Properly", "Let’s solve for $ z $ such that:", "1. $ \cos z = 0 $\n2. $ \sin z = \frac{1}{2} $", "From (1), $ z = \frac{\pi}{2} + k\pi $.\nSubstitute into (2):", "- For $ k $ even: $ z = \frac{\pi}{2} + 2m\pi $, then $ \sin z = 1 $\n- For $ k $ odd: $ z = \frac{3\pi}{2} + 2m\pi $, then $ \sin z = -1 $", "Neither equals $ \frac{1}{2} $, proving that no such $ z $ exists.", "### Alternative Interpretation: Values of $ z $ Where $\cos z = 0$, But $\sin z <br/>\ne \frac{1}{2}$", "While there is no angle where both conditions are satisfied, the set $ \cos z = 0 $ still defines crucial solutions in complex analysis and periodic functions — particularly when solving equations like:", "[\n\cos z = 0 \quad \Rightarrow \quad z = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z})\n]", "Each of these angles corresponds to a peak or trough of sine waves, highlighting the orthogonality central to trigonometry.", "### Conclusion", "While $ \cos z = 0 $ clearly at $ z = 90^\circ, 270^\circ $, the sine evaluates to $ \pm 1 $, never $ \frac{1}{2} $. Hence, there are no solutions satisfying both $ \cos z = 0 $ and $ \sin z = \frac{1}{2} $. This distinction reinforces the foundational principle in trigonometry: at angles where cosine vanishes, sine takes extreme values—positive or negative unity—but never intermediate values such as $ \frac{1}{2} $.", "For deeper understanding, explore the unit circle and phase relationships in periodic functions, where co-function identities and symmetry reveal the relationships between sine and cosine across the complex plane.", "---", "Keywords: $\cos z = 0$, $\sin z = \frac{1}{2}$, complex trigonometry, unit circle, periodic functions, $z$-values, trigonometric identities, sine and cosine properties.", "Meta Description: Discover why there are no angles $ z $ such that $ \cos z = 0 $ and $ \sin z = \frac{1}{2} $. Learn how trigonometric identities and the unit circle clarify these fundamental relationships."]