# Sine Function

## What is Sine Function?

The sine function is a **periodic function** which is very important in trigonometry.

The simplest way to understand the sine function is to use the unit circle. In trigonometry, a **unit circle** is the circle of radius **one** centered at the origin \(\left( {0,0} \right)\) in the Cartesian coordinate system.

Now that we have seen how to define trigonometric ratios for arbitrary angles, we are ready to study trigonometric ratios from the perspective of functions.

Consider the following function definition:

\[f\left( x \right) = \sin x\]

The argument \(x\) is any real number. When we apply the **sine function** to \(x\), the output generated by \(f\) will be the sine of \(x\) radians. For example,

\[\begin{align}&f\left( 0 \right) = \sin 0 = 0\\&f\left( {\frac{\pi }{3}} \right) = \sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}\\&f\left( {\frac{{5\pi }}{6}} \right) = \sin \frac{{5\pi }}{6} = \frac{1}{2}\\&f\left( {1000\pi } \right) = \sin 1000\pi = 0\end{align}\]

**✍****Note:** Sine Function can be applied to every real number, because as we have discussed in conversion relations of trigonometric ratios, we can calculate the **sine** of any real number.

## Graph of Sine Function:

**Think:** What will the graph of the sine function look like? To understand that, let us analyze in more detail the variation in the value of the output of the sine function, as the input (the angle) varies.

The following figure shows a **unit circle** with center \(O\), and a point \(P\) moving along the circumference of this circle. The angle which \(OP\) makes with the positive direction of the \(x\)-axis is \(x\) (radians):

\(PQ\) is the perpendicular dropped from \(P\) to the horizontal axis. We note that

\[\sin x = \frac{{PQ}}{{OP}} = \frac{{PQ}}{1} = PQ\]

As \(x\) varies, we can study the variation in \(\sin x\) as the variation in the length of \(PQ\).

Suppose that initially, \(P\) is on the horizontal axis. Let us consider a movement of \(P\) through \(90^\circ \) or \(\frac{\pi }{2}\) rad. The following figure shows different positions of \(P\) for this movement:

Clearly, \(PQ\) has increased in length, from an initial value of 0 (when \(x\) is 0 radians) to a final value of 1 (when \(x\) is \(\frac{\pi }{2}\) radians).

We can now plot this variation. We use the horizontal axis for \(x\) (the input variable, the angle in radians), and the vertical axis for the sine of \(x\) (which is the same as the length \(PQ\)). We obtain a plot shown below (this is a computer-generated plot, so the shape is pretty accurate):

Now, let us see what happens as \(P\) moves further. The following figure shows different positions of \(P\) as it subsequently moves from a \(90^\circ \) position to \(180^\circ \) position:

In this phase of the movement, the length \(PQ\) decreases, from a maximum of 1 at \(90^\circ \), to a minimum of 0 at \(180^\circ \). We continue plotting this variation on the same graph we plotted earlier:

When \(P\) moves from a position of \(180^\circ \) to a position of \(270^\circ \), *PQ* decreases from 0 to \( - 1\) (though the length or magnitude of \(PQ\) increases, since the direction is negative-*y*, the **actual algebraic value** of \(PQ\)*,* or the sine of the angle, decreases):

We add this variation to our graph:

Finally, when \(P\) moves from a position of \(180^\circ \) to a position of \(360^\circ \), \(PQ\) increases from \( - 1\) to 0 (once again, the length or magnitude of \(PQ\) decreases, but the algebraic value of \(PQ\), or the sine of \(x\), increases):

Adding this variation, we obtain the complete plot of \(PQ\) versus \(x\), or \(\sin x\) versus \(x\), for one complete cycle of 0 radians to \(2\pi \) radians (\(0^\circ \) to \(360^\circ \)):

What happens when \(P\) moves further now? The same variation cycle starts all over again. Thus, if we extend the sine function to take on all real inputs, we obtain the following graph:

Note the periodic nature of this graph. If \(P\) moves through a cycle of \(2\pi \) radians, it will come back to its original position, so the sine value will return to its original value. We can write this fact as follows:

\[\sin \left( {x + 2\pi } \right) = \sin x\,\,\,{\rm{for\,\,\, every}}\,x\]

This means that no matter what \(x\) is initially if you add \(2\pi \) to it, you come back in the **same direction**, and so the sines of the two angles are the same. In general, if you add any multiple of \(2\pi \) to \(x\), you come back to the same direction, and so we can say that:

\[\sin \left( {x + 2n\pi } \right) = \sin x\,\,\,{\rm{for \,\,\,every}}\,x\]

For example,

\[\begin{align}&\sin \left( {1000\pi + \frac{\pi }{3}} \right) = \sin \frac{\pi }{3}\\&\sin \left( {\frac{{2\pi }}{3} - 128\pi } \right) = \sin \frac{{2\pi }}{3}\end{align}\]

**✍****Note:** The domain of the sine function is \(\mathbb{R}\), and the range is \(\left[ { - 1,1} \right]\).

## Solved Example:

**Example 1:** For what values of \(x\) is \(\sin x = \frac{1}{2}\)?

**Solution:** When we were restricted to considering only angles between \(0^\circ \) and \(90^\circ \), we would have given an answer of \(30^\circ \) or \(\frac{\pi }{6}\) radians for this question. However, now we must think of the sine term as a function. And since this function is periodic, there will be infinitely many values of \(x\) for which \(\sin x = \frac{1}{2}\). If we draw the line \(y = \frac{1}{2}\) on the graph of \(y = \sin x\), the *x*-coordinates of all the points of intersection will satisfy the given equation:

In the interval \(\left[ {0,2\pi } \right]\), we see that the two solutions for \(\sin x = \frac{1}{2}\) are \(x = \frac{\pi }{6}\) and \(x = \frac{{5\pi }}{6}\). Now, all we need to do is to take the corresponding solutions in each subsequent (and preceding) interval spanning \(2\pi \). For example, in the interval \(2\pi \) to \(4\pi \), the solutions to \(\sin x = \frac{1}{2}\) will :

\[\begin{align}&x = 2\pi + \frac{\pi }{6} = \frac{{13\pi }}{6}\\&x = 2\pi + \frac{{5\pi }}{6} = \frac{{17\pi }}{6}\end{align}\]

How can we specify the entire solution set? In each interval of the form \(\left[ {2n\pi ,\left( {2n + 2} \right)\pi } \right]\), one solution will be \(x = 2n\pi + \frac{\pi }{6}\), and the other solution will be \(x = 2n\pi + \frac{{5\pi }}{6}\). Thus, we can write the solution set as follows:

\[\boxed{x = \left\{ {2n\pi + \frac{\pi }{6},2n\pi + \frac{{5\pi }}{6}} \right\},\,\,\,n \in \mathbb{Z}}\]

**Challenge:** Write the solution set for \(\sin x = 1\)?

**⚡Tip:** Use a similar approach as in the above example.