Bhaskara l's Approximation $$ \sin x \approx \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} $$

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1. We start with Bhaskara I’s sine approximation formula defined for $$ 0\le x\le \pi $$:
   
   $$
   \sin x \approx \frac{16\,x\,(\pi - x)}{5\pi^2-4\,x\,(\pi - x)}
   $$

2. Check the behavior at the endpoints and at the maximum:

- At $$ x=0 $$:
     
     The numerator is 
     
     $$
     16\cdot0\cdot(\pi-0)=0,
     $$
     
     and the denominator is 
     
     $$
     5\pi^2-4\cdot0\cdot(\pi-0)=5\pi^2.
     $$
     
     Thus, 
     
     $$
     \sin 0 \approx \frac{0}{5\pi^2}=0.
     $$

- At $$ x=\pi $$:
     
     The numerator is 
     
     $$
     16\cdot\pi\cdot(\pi-\pi)=0,
     $$
     
     and the denominator is 
     
     $$
     5\pi^2-4\cdot\pi\cdot (\pi-\pi)=5\pi^2.
     $$
     
     Hence, 
     
     $$
     \sin \pi \approx \frac{0}{5\pi^2}=0.
     $$

- At $$ x=\frac{\pi}{2} $$:
     
     We compute:
     
     $$
     \text{Numerator} = 16\cdot\frac{\pi}{2}\cdot\left(\pi-\frac{\pi}{2}\right)=16\cdot\frac{\pi}{2}\cdot\frac{\pi}{2} = 16\frac{\pi^2}{4}=4\pi^2.
     $$
     
     Also,
     
     $$
     \text{Denominator} = 5\pi^2-4\cdot\frac{\pi}{2}\cdot\frac{\pi}{2}=5\pi^2-4\frac{\pi^2}{4}=5\pi^2-\pi^2=4\pi^2.
     $$
     
     Therefore,
     
     $$
     \sin\frac{\pi}{2} \approx \frac{4\pi^2}{4\pi^2}=1.
     $$

3. Notice that the formula is symmetric with respect to $$ x $$ and $$ \pi-x $$. This symmetry reflects the property of the sine function in the interval $$ [0,\pi] $$, since

$$
   \sin x = \sin (\pi-x).
   $$

4. The approximation is constructed so that its value, as well as its qualitative behavior (vanishing at $$ 0 $$ and $$ \pi $$ and peaking at $$ \frac{\pi}{2} $$), mimic those of the true sine function. Although it is not an exact equality for all $$ x $$, Bhaskara I’s formula provides a very good approximation for practical purposes within the given interval.

5. In summary, Bhaskara I's approximation

$$
   \sin x \approx \frac{16\,x\,(\pi-x)}{5\pi^2-4\,x\,(\pi-x)}
   $$

reproduces the key properties of the sine function over $$ 0\le x\le \pi $$. It satisfies $$\sin 0=0$$, $$\sin\frac{\pi}{2}=1$$, and $$\sin\pi=0$$, while maintaining the essential symmetry $$ \sin x=\sin (\pi-x) $$.
Bhaskara I's sine approximation formula is: $$ \sin x \approx \frac{16x(\pi - x)}{5\pi^2 - 4x(\pi - x)} $$ This formula accurately represents the sine function for $$ 0 \leq x \leq \pi $$, matching the values at $$ x = 0 $$, $$ x = \frac{\pi}{2} $$, and $$ x = \pi $$. It also maintains the symmetry $$ \sin x = \sin (\pi - x) $$, making it a reliable approximation within the specified interval.

Bhaskara l's Approximation \( \sin x \approx \frac{16 x(\pi-x)}{5 \pi^{2}-4 x(\pi-x)} \)

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