An odd function is symmetrical about the origin. In other words, if \((a, b)\) is a point on the function, then \((-a, -b)\) is also a point on the function, assuming that both \(a\) and \(-a\) are in the domain of the function. If this function, \(f\), is an odd function, then its property can also be written as this equation:
$$f(-x) = -f(x)$$
The sine function is one such odd function. That is, for any real \(x\):
$$\sin(-x) = -\sin(x)$$
Example 1: The property above holds for \(x = \frac{\pi}{6}\):
$$\sin \left(-\frac{\pi}{6}\right) = -\sin \left(\frac{\pi}{6}\right) \Rightarrow$$
$$\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}$$
This is true. If it benefits your class, show them the relevant points on a Unit Circle.
Example 2: Although rather trivial, \(x\) can equal \(0\):
$$\sin(-0) = -\sin(0) \Rightarrow$$
$$\sin(0) = -0 \Rightarrow$$
$$0 = 0$$
Example 3: The argument inside the sine function is a function in itself, perhaps more complicated than just \(x\). Let this inside function be \(g(x)\). So we want to prove that
$$\sin(-g(x)) = -\sin(g(x))$$
Well it turns out the argument inside the sine function does not interfere with the sine function being odd, so this property holds regardless. Thus
$$\sin(-g(x)) = -\sin(g(x))$$
Consider showing this being true for a specific example for a specific \(g(x)\) other than \(-x\).
Example 4: If a constant is added to the sine function, the sum of these two functions is no longer odd. For instance, we can see that the following is NOT true if \(f(x) = \sin(-x) + 1\):
$$\sin(-x) + 1 = -(\sin(x) + 1)$$
To see why this is not true, distribute the negative sign on the right side:
$$\sin(-x) + 1 \neq -\sin(x) - 1$$
Add \(1\) to both sides of the equation:
$$\sin(-x) + 2 \neq -\sin(x)$$
The sine function by itself is still odd, thus \(\sin(-x) = -\sin(x)\), so we can substitute this on the left side of the equation:
$$-\sin(x) + 2 \neq -\sin(x)$$
Add \(\sin(x)\) to both sides of the equation:
$$2 \neq 0$$
Of course these numbers aren't equal to each other! In fact, any constant can be substituted for the \(1\) and the two forms will not be equal.