Prove that sine function is continuous at every real number.

To prove that the sine function is continuous at every real number, we need to show that it satisfies the epsilon-delta definition of continuity. According to this definition, for any epsilon greater than zero, there exists a delta greater than zero such that the absolute difference between the sine of two real numbers x and y is less than epsilon whenever the absolute difference between x and y is less than delta.

Let's proceed with the proof:

Proof:
We want to prove that for any given epsilon > 0, there exists a delta > 0 such that for all x and y satisfying |x - y| < delta, we have |sin(x) - sin(y)| < epsilon.

Using the trigonometric identity, we have:
sin(x) - sin(y) = 2 * cos((x + y)/2) * sin((x - y)/2).

Since cos((x + y)/2) and sin((x - y)/2) are continuous functions, we can focus on them separately.

Continuity of cos((x + y)/2):
The function cos((x + y)/2) is the composition of the continuous function cos(z) with z = (x + y)/2. Hence, cos((x + y)/2) is continuous.

Continuity of sin((x - y)/2):
Similarly, the function sin((x - y)/2) is the composition of the continuous function sin(z) with z = (x - y)/2. Hence, sin((x - y)/2) is continuous.

Since both cos((x + y)/2) and sin((x - y)/2) are continuous, their product 2 * cos((x + y)/2) * sin((x - y)/2) is also continuous.

Now, let's choose delta = epsilon. If |x - y| < delta, then we have:

|x - y| < epsilon.

Using the triangle inequality, we can write:

|sin(x) - sin(y)| = |2 * cos((x + y)/2) * sin((x - y)/2)| ≤ 2 * |cos((x + y)/2)| * |sin((x - y)/2)|.

Since both cos((x + y)/2) and sin((x - y)/2) are bounded by 1 in absolute value, we have:

|sin(x) - sin(y)| ≤ 2 * |cos((x + y)/2)| * |sin((x - y)/2)| ≤ 2 * 1 * 1 = 2.

Therefore, |sin(x) - sin(y)| < epsilon whenever |x - y| < epsilon.

This completes the proof, showing that for any epsilon > 0, there exists a delta > 0 such that |sin(x) - sin(y)| < epsilon whenever |x - y| < delta. Hence, the sine function is continuous at every real number.