Actually, the **maximum finite value of $r$** occurs when $\sin(2\theta)$ is at its **maximum**, because $r^2 = \frac{100}{\sin(2\theta)}$, and $\sin(2\theta) \leq 1$.

["Understanding the Maximum Finite Value of $ r $: How $ \sin(2\ heta) $ Plays a Key Role", "When analyzing polar equations of the form $ r^2 = \frac{100}{\sin(2\ heta)} $, a fundamental concept emerges about the finite maximum value of $ r $. To fully grasp this, we must explore the behavior of the trigonometric function $ \sin(2\ heta) $ and its impact on the expression.", "### $ r^2 = \frac{100}{\sin(2\ heta)} $: The Core Relationship", "For real and finite $ r $ values, the denominator $ \sin(2\ heta) $ must remain strictly positive and less than or equal to 1. Since $ r > 0 $, the equation $ r^2 = \frac{100}{\sin(2\ heta)} $ implies:", "$$\nr = \frac{10}{\sqrt{\sin(2\ heta)}}\n$$", "To maximize $ r $, we must minimize $ \sqrt{\sin(2\ heta)} $, which in turn demands minimizing $ \sin(2\ heta) $ — while keeping it positive.", "### The Maximum of $ \sin(2\ heta) $ Determines the Maximum $ r $", "Recall that the sine function satisfies $ \sin(x) \leq 1 $ for all real $ x $. Therefore, the maximum finite value of $ \sin(2\ heta) $ is 1. This occurs when:", "$$\n\sin(2\ heta) = 1 \quad \Rightarrow \quad 2\ heta = 90^\circ + 360^\circ n \quad (n \in \mathbb{Z}) \quad \Rightarrow \quad \ heta = 45^\circ + 180^\circ n\n$$", "Substituting $ \sin(2\ heta) = 1 $ into the equation for $ r^2 $:", "$$\nr^2 = \frac{100}{1} = 100 \quad \Rightarrow \quad r = \sqrt{100} = 10\n$$", "Thus, the maximum finite value of $ r $ is 10, achieved precisely when $ \sin(2\ heta) $ reaches its peak:\n$$\n\boxed{ \ ext{Maximum finite } r = 10 \ ext{ when } \sin(2\ heta) = 1 }\n$$", "### Why This Matters", "This principle is critical in polar coordinate geometry and has practical implications in physics, engineering, and computer graphics. Recognizing how $ \sin(2\ heta) $ governs $ r $ helps in plotting curves, optimizing distances, or solving for coordinates in complex curves — always bearing in mind that $ r $ remains finite only when the denominator $ \sin(2\ heta) $ is positive and bounded away from zero.", "### Summary", "- The expression $ r^2 = \frac{100}{\sin(2\ heta)} $ yields maximum finite $ r $ when $ \sin(2\ heta) = 1 $\n- $ \sin(2\ heta) \leq 1 $, so $ r \leq 10 $, confirming that 10 is the upper finite limit\n- This insight supports accurate interpretation and manipulation of polar equations", "Embrace the power of trigonometric maxima — they control the reach of $ r $ in polar landscapes.", "---", "### SEO Keywords:\n- maximum finite value of $ r $\n- polar coordinates $ r $ and $ \ heta $\n- $ \sin(2\ heta) maximum $\n- polar equation $ r^2 = \frac{100}{\sin(2\ heta)} $\n- how $ \sin(2\ heta) $ limits $ r $\n- polar curve $ r^2 = \frac{100}{\sin(2\ heta)} $\n- maximum $ r $ in polar equations", "Optimize your mathematics comprehension with the key insight: maximum finite $ r $ occurs when $ \sin(2\ heta) $ is at its maximum value of 1, making $ r = 10 $."]