We develop here the details required for a good understanding of the sine function.  Once you understand the material presented here, the cosine function can be easily understood through an analogous development.

The quantity A sin(`theta) is defined to be the y coordinate of the point whose angular position, on a circle of radius A whose center is at the origin of an x-y coordinate system, is `theta, as measured from the point that intercepts the positive x axis.

The graph we obtain when we plot the y coordinate of this point, with a circle positioned with its center on the horizontal axis, is therefore the graph of y = A sin(`theta).

We can interpret this graph as a position vs. clock time graph, or as a displacement vs. position graph:

• If `theta represents the angular position at clock time t corresponding to motion with angular velocity `omega, then `theta = `omega * t and our graph represents y = A sin(`omega * t) vs. t.
• If `theta = k * x, where x is a position along a wave and k is a constant (called the wave number), then our graph could represent y = A sin(k * x) vs. x. This graph would represent a 'snapshot' of the actual wave.

We can find the exact value of the sine of any multiple of `pi/6 or `pi/4 from basic geometry, using the Pythagorean Theorem and very basic facts about triangles.

• If we sketch on a circle of radius A the sector whose angle is `pi/6, as depicted below, we can form the indicated right triangle, with y as the side opposite the angle.
• We are familiar with the fact that `pi / 6 = 30 degrees. For a moment it will be simpler to work in degrees.
• If we place two copies of this right triangle with as shown below, we form an equilateral triangle whose sides are A.
• It should be clear from this figure that y = A / 2.
• Since y = A sin(`pi / 6), we have A sin(`pi/6) = A / 2. We can solve this equation to obtain sin(`pi / 6) = 1/2.

We can also find the sine of `pi/4, or 45 deg, by looking at the triangle formed as indicated in the figure below.

• From the symmetry of the triangle we can see that the two shorter sides of the triangle are equal. We call the length of the sides s.
• From the Pythagorean Theorem we therefore see that A^2 = s^2 + s^2.
• Solving this equation for s, as indicated below, we see that s = 1 / `sqrt(2) * A.
• Setting y = A sin(`pi / 4) = s = 1/`sqrt(2) * A, we have an equation that we can solve for sin(`pi/4).
• Solving the equation we see that sin(`pi/4) = 1/`sqrt(2) = `sqrt(2) / 2.

The approximate values of `sqrt(2) / 2 and `sqrt(3) / 2 are .71 and .87.

The graph is shown below in some detail. The dotted lines are y = 1/2, y = .71, y = .87 and y = 1.

• We note that `pi/6, `pi/3 and `pi/2 are equally spaced, dividing the interval from 0 to `pi/2 into 3 equal segments (not shown too accurately on the figure below; you should be able to do better).
• `pi/4 cuts the interval from 0 to `pi/2 in half, as depicted below; `pi/4 also lies halfway between `pi/6 and `pi/3.

We can extend our table past `pi/2 to encompass angular positions around the entire circle (and beyond, into a second or third cycle, actually into any number of cycles, if we wish).

• The table below lists values of sin(`theta) vs. `theta for multiples of `theta = `pi/6, up through the third-quadrant angle 4 `pi / 3.
• The table is made with the aid of the circle depicted at right.
• The symmetry of the situation dictates the indicated y coordinates on the circle, in an obvious way.
• You should be able to produce this table and this picture instantly anytime you need it.

In the next figure the horizontal dotted lines depict the y values corresponding to multiples of `pi/6.

• The vertical dotted lines divide the cycle from 0 to `pi/2 into 12 equal intervals, corresponding to the multiples of `pi/6.
• You should be able to produce this graph instantly anytime you need it.