Physics Etc: Sine & Cosine

Sine and Cosine Functions: Derivatives, and Taylor Series

Triangles were introduced in the second article, Ancient Functions, where we learned that the interior angles of any triangle sum to 180°, and we studied the right triangle through some of its properties. To recap, the right triangle has one side called the hypotenuse, being positioned opposite to the right angle, and also two shorter sides, the lengths of which are related by the Pythagorean theorem. We will now add the sine and cosine functions to our trigonometry. The sine function is defined as the ratio of the length of the side opposite the (acute) angle to the length of the hypotenuse (adjacent to the acute angles). And, the cosine function is defined as the ratio of the length of the side adjacent to the acute angle to the length of the hypotenuse.

\begin{equation}\label{sohcahtoa} \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \end{equation}

We know a right triangle is one that has a 90° angle, and so the sum of the other two interior angles must be 90°. First, let's derive the Triangle Area Formula using the sine function and some geometry.

The Triangle Area Formula (TAF) states that the area of a triangle, $A_\triangle$, is equal to one-half the product of the two adjacent side lengths, $a$ and $b$, times the sine of the included angle, $\theta$. The proof uses the base-side being arbitrary, so if it works for one side of a general triangle, it will work for any angle of .

\begin{equation}\label{TAF} A_\triangle = \frac{1}{2} ab \sin(\theta) \end{equation}

Which is immediately proven for the right triangle, where $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c}$, and the area is $A_\triangle = \frac{1}{2}ab = \frac{1}{2}bc\sin(\theta)$.

In general, a triangle does not have a right angle, which is why the TAF is useful for finding its area. Using an overlay to a general triangle with arbitrary orientation, the TAF is directly proven with an overlay of right triangles, which will demonstrate the decomposition into right triangles, so we can use the tools of right angles.

Figure 1. Triangle Area Formula Illustration, here is a diagram of an arbitrary triangle, $\triangle{}OQP$, in red, oriented with the obtuse angle at the bottom left, and the side $\overline{OQ}$ is horizontal and located at the bottom. The acute angle of interest is $\angle{}OQP =\angle{}OQR = \theta$, and there are two overlaid right triangles, $\triangle{}ORQ$ and $\triangle{}PQR$.

$$ \theta = \angle{}OQP = \angle{}OQR$$
$$\overline{OR} = \overline{OQ}\sin(\theta) $$
$$\overline{UP} = \overline{PQ}\sin(\theta) $$

The area of the primary triangle, $\triangle{}OQP$, is equal to the sum of the areas of the two inscribed right triangles with shared side, $\overline{OR}$, $\triangle{}ORQ$ and $\triangle{}POR$. The area of the right triangle $\triangle{}ORQ$ is equal to one-half the product of its two shorter sides, $\overline{OR}$, and its height, $\overline{RQ}$, which is $A_{\triangle{}ORQ} = \frac{1}{2} \overline{OR}\, \overline{RQ}$. The area of the right triangle $\triangle{}POR$ is equal to one-half the product of its base,$\overline{PR}$ , and its height, $\overline{OR}$, which is $A_{\triangle{}POR} = \frac{1}{2} \overline{OR}\, \overline{PR}$. So the area of the triangle $\triangle{}OQP$ is equal to the sum of these two areas, $A_{\triangle{}OQP} = \frac{1}{2} \overline{OR}\, \overline{RQ} + \frac{1}{2} \overline{OR}\, \overline{PR}$, or simplifying, $A_{\triangle{}OQP} = \frac{1}{2} \overline{OR}(\overline{RQ} + \overline{PR})=\frac{1}{2} \overline{OR}\, \overline{PQ}$. And substituting for $\overline{OR}$, we have $A_{\triangle{}OQP} = \frac{1}{2} \overline{OQ}\, \overline{PQ}\sin(\theta)$, which is the TAF as stated in \eqref{TAF}.

$$ A_{\triangle{}UPQ} = A_{\triangle{}OPQ} + A_{\triangle{}UPO} $$

To consider the other choice of the hypotenuse for angle $\theta$, we consider the triangle $\triangle{}UPQ$, where angle $\angle{}PUQ = 90^\circ$ as indicated with the small square at vertex $U$. Now the side $\overline{PQ}$ is the hypotenuse for angle $\theta$, and the side $\overline{UQ}$ is the short side adjacent to angle $\theta$, to use the right angle triangle area formula. So the area of triangle $\triangle{}UPQ$ is equal to the sum of the areas of triangles $\triangle{}OPQ$ and $\triangle{}UPO$, or rearranging, $A_{\triangle{}OPQ} = A_{\triangle{}UPQ} - A_{\triangle{}UPO} $, which is the TAF as stated in \eqref{TAF} since $\overline{OQ} = \overline{UQ} - \overline{UO}$, or:

$$ A_{\triangle{}OPQ} = A_{\triangle{}UPQ} - A_{\triangle{}UPO} = \frac{1}{2} \overline{UQ}\, \overline{PQ}\sin(\theta) - \frac{1}{2} \overline{UO}\, \overline{PQ}\sin(\theta) = \frac{1}{2} \overline{OQ}\, \overline{PQ}\sin(\theta)$$.

Which is what we set out to prove.

The Unit Circle

The unit circle is often used to define the trigonometric functions, such as sine, cosine, and tangent, which are based on the coordinates of points on the circle. The unit circle is a fundamental concept in trigonometry and geometry, and it is defined as a circle with a radius of one unit, centered at the origin of a Cartesian coordinate system, where a point on the circle is identified with an angle, $\angle POQ$ in the diagram below in figure 2. The unit circle can be considered as a set of points with coordinates $(x,y)$ that satisfy the equation $x^2 + y^2 = r^2=1$, where $r$ is the hypotenuse which is now configured to be a radius.

The unit circle is also used to visualize the relationships between the trigonometric functions and the angles they represent. In the unit circle, an angle is measured in radians, which is a unit of measure that is based on the radius of the circle. One radian is defined as the angle that subtends an arc length equal to the radius of the circle. Since the circumference of a circle is $2\pi$ times its radius, where $\pi$ is a constant convention for the ratio of the circumference to the diameter of a circle. There are $2\pi$ radians in a full circle, which is equivalent to 360 degrees.

Figure 2. Unit Circle Illustration, here is a diagram of the unit circle, centered at the origin, $O$, of a Cartesian coordinate system. The angle $\theta$ is shown in standard position, with the hypotenuse (radius) extending from the origin to the point $(x,y)$ on the circle.

Sine and cosine are very similar to one another, called cofunctions, and they have characteristics born of the right triangle properties. The Angle Addition Formulas for the two cofunctions, are proven most directly using Euler's Formula; this will demonstrate that they are related by a phase difference of $90^\circ$—which is what makes them cofunctions. The cofunctions are ratios of a side (opposite or adjacent) to the hypotenuse, as functions of an acute angle of a right triangle.

SOH-CAH-TOA is the second most important thing in trigonometry (after Pythagoras, of course), which gives the relationships of the sides of the right angle triangle to the three (very common) functions, sine, cosine, and tangent in a three-syllable mnemonic.

Area of a circle: $$ \pi r^2 $$ where $r$ stands for the radius, or distance from the center of the circle to the edge. And note that the area of the square which is $2r$ on the side, is $4r^2$ and that the circle covers roughly 3/4 of the square. Pi was first studied by Archimedes, who bounded the number $\pi$ by the sum of sides of a smaller polygon (inscribed) and a larger one (circumscribed). So the circumferance of the circle ($2\pi r$, or $\pi$ times the diameter) is between the inner polygon side length (less than) and the outer polygon side length (greater than). $$ Nl_{inner} \lt 2\pi r \lt Nl_{outer} $$ where $l$ is the distance between adjacent polygon vertices, and $N$ is the number of sides (Archimedes reportedly made this calculation for $N=96$).

The natural angle unit is the radian, which is defined as the angle subtending an arc length equal to the radius of the circle. Since the circumference of a circle is $2\pi r$, there are $2\pi$ radians in a full circle, which is approximately $6.28$.

The Taylor Series

To calculate the Taylor series, we first calculate the derivative of the sine function, which by plotting the two functions side by side, it can be seen that for any point the slope of the sine function is equal to value of the cosine function. And, that for any point of the cosine function, its slope is equal to negative the value of the sine function.

The derivative of sin(x) evaluated at $\pi/2$ is zero, and periodically, but its not zero everywhere, so it's not a zero function.

Figure 3. Sine and Cosine Functions, here are the sine function in red, and the cosine function in green, plotted over the domain from -10 to 10. The sine function has zeros at integer multiples of $\pi$, and the cosine function has zeros at odd integer multiples of $\frac{\pi}{2}$.

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