Cofunction Series

The right triangle forms the basis for the sine and cosine functions to our trigonometry. The sine function is defined as the ratio of the length of the side opposite of the acute angle to the length of the hypotenuse. The cosine function is defined as the ratio of the length of the short side adjacent to the length of the hypotenuse, or the complementary angle's sine. And the tangent is the slope of the hypotenuse when the adjacent is parallel to the $x$-axis, that is the ratio of opposite to adjacent lengths.

Let's derive the Sine Rule for Area using the sine function and Euclidean geometry.

The Sine Rule for Area (SRA) states that the area of a triangle, $A_\triangle$, is equal to one-half the product of two adjacent side lengths, $a$ and $b$, times the sine of the included angle, $\theta$.

In general, a triangle does not have a right angle, which is why the SRA is useful for finding its area as it introduces a cofunction to the . Using an overlay of inscribed right triangles, using the fact that you can always compose any triangle using inscribed right triangles, we can then use the tools of right angles.

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

Where the SRA variables, $a$ and $b$, are represented by the lengths of the sides $\overline{OQ}$ and $\overline{PQ}$, and the angle $\theta=\angle{}OQP$ is the included angle between those two sides.

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 SRA as stated in \eqref{SRA}.

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

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 SRA as stated in \eqref{SRA} 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, 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.

Points on the circle are naturally described in terms of the radius, $r=1$ here, and angle, $\theta$ extending into the first quadrant for the $\theta=28.8^\circ=0.502$ shown here, which is the plane polar coordinate system, where $\Phi:(\theta, r)\mapsto (x,y)$ is the atlas from polar coordinates to cartesian coordinates.

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, where the tangenet is opposite-over-adjacent.

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 Cofunction 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 proof of the derivative uses the area of a circle, the angle addition formula of sine, and the triangle area formula.

Derivative of Sine

$$ \begin{equation}\label{sine_derivative} \frac{d}{dx}\sin{(x)}=\lim_{dx\to 0}\frac{\sin{(x + dx)}-\sin{(x)}}{dx} \end{equation} $$

The sine term with the addition of the two angles is simplified with the angle addition formula, derived from Euler's Formula with angle addition and the use of the exponential identity:

$$ e^{i(\theta + \phi)}=e^{i\theta}e^{i\phi} $$

We will also use the linear independence of the pure real and pure imaginary components in the equation.

$$ \cos{(\theta + \phi)} + i\sin{(\theta + \phi)} = (\cos{(\theta)} + i\sin{(\theta )})(\cos{(\phi)} + i\sin{(\phi)}) $$

Factoring out terms and separating real and imaginary:

$$ = \cos{(\theta)}\cos{(\phi)} - \sin{(\theta )}\sin{(\phi )} + i(\sin{(\theta )}\cos{(\phi)} +\cos{(\theta)}\sin{(\phi)}) $$

Yields:

$$ \sin(\theta + \phi) = \sin{(\theta )}\cos{(\phi)} +\cos{(\theta)}\sin{(\phi)} $$

Applying to the derivative, \eqref{sine_derivative}:

$$ \frac{d}{dx}\sin{(x)}=\lim_{dx\to 0}\frac{\sin{(x )}\cos{(dx)} +\cos{(x)}\sin{(dx)}-\sin{(x)}}{dx} $$

Next we use approximations for a small angle of the cofunctions appearing here.

Proof $\lim_{dx\to 0}\sin{(dx)}=dx$

The area of the unit circle, being $A=\pi r^2 = \pi$, is sectioned by the inscribed angle so that the area of its wedge is the angle (in radians) over two, as the fraction of the circle times the area of the circle has an equal number of the constant, $\pi$, in the numerator as the denominator. For the area subtended by the inscribed small angle, $dx$, it is exactly $\frac{dx}{2\pi}A=\frac{dx}{2}$. This area is approximately equal to the area of the triangle OQP, with sides of length exactly one, and inscribed angle, $dx$, $A_{\triangle_{OQP}}=\frac{1}{2}\sin{(dx)}$. Setting these two areas as approximately equal yields the desired result.

Cosine of a small angle is obtained from the Pythagorean theorem on the inscribed right triangle, and using $\sin{(dx)}\approx dx$:

$$ \cos^2{(dx)} + \sin^2{(dx)} = 1 $$

Solving for $\lim_{dx\to 0}\cos{(dx)}$

$$ \cos{(dx)} = \sqrt{1-dx^2} $$

Now this is a normalized binomial form with power one-half, so we can use the Binomial Approximation (lemma of the Binomial Theorem):

$$ \cos{(dx)} \approx 1-\frac{1}{2}dx^2 $$

So since second order appearances of the infinitesimal quantity, $dx$, go to zero in the limit of the derivative, we can leave out the second order terms, and just keep unity.

Therefore we see the derivative of sine is to first order in the infinitesimal:

$$ \frac{d}{dx}\sin{(x)}=\lim_{dx\to 0}\frac{\cos{(x)}dx}{dx} = \cos{(x)} $$

The derivative of cosine uses the angle addition formula for cosine, and the rest is the same, yielding:

$$ \frac{d}{dx}\cos{(x)}=-\sin{(x)} $$

Taylor Series Constructed

The Taylor Series around zero is the Maclarin Series, and for Sine and Cosine, they are as follows:

$$ \sin(x) = \sin(0) + \cos(0) x - \frac{1}{2!}\sin(0) x^2 - \frac{1}{3!}\cos(0) x^3 + \cdots $$

Here we see that the sine function is to first order linear in its argument—applicable when $x\ll 1$, which we knew from geometric argument, above, is now reflected in the Taylor series, which has more terms of precision built-in.

$$ \cos(x) = \cos(0) - \sin(0) x - \frac{1}{2!}\cos(0) x^2 + \frac{1}{3!}\sin(0) x^3 + \cdots $$

Tangent Taylor Series

As a rational construction of the cofunctions, the tangent has no dependence on the hypotenuse as the Sine in the numerator and Cosine in the denominator have the same factor of it which forms a factor of unity upon reduction. The derivatives of tangent rely on the derivatives of the cofunctions (via the Product Rule here) starting procedurally with the definition of the derivative.

Derivative of Tangent

$$ \frac{d}{dx}\tan(x) = \lim_{dx\to 0}\frac{tan(x+dx)-tan(x)}{dx} $$

Using the angle addition rule on the tangent is a matter of using the angle addition formulas for the cofunctions:

$$ \tan(x+dx)=\frac{\sin(x)\cos(dx)+\sin(dx)\cos(x)}{\cos(x)\cos(dx) -\sin(x)\sin(dx)} $$

Applying our small angle approximations for the cofunctions in order to get first-order approximations for terms with argument of $dx$, and plugging into the numerator:

$$ \tan(x+dx)-\tan(x)=\frac{\sin(x)+dx\cos(x)}{\cos(x) -\sin(x)dx} - \frac{\sin(x)}{\cos(x)} $$

Finding equivalent fractions with a common denominator, and reducing:

$$ \frac{d}{dx}\tan(x) = \lim_{dx\to 0}\frac{1}{\cos^2(x)-\sin(x)\cos(x)dx}=\cos^{-2}(x) = \sec^2(x) $$

Second Derivative of Tangent

Applying the Chain Rule to the secant-squared function gives us:

$$ \frac{d^2}{dx^2}\tan(x)=\frac{d}{dx}\cos^{-2}(x)=\frac{2\sin(x)}{\cos^3(x)}=2\sec^2(x)\tan(x) $$

Constructing the Tangent Taylor Series

We know what the zeroth derivative of tangent is at zero, $\tan(0)=0$, and its first derivative at zero, $\sec^2(0)=1$, and calcululating the derivatives through the fifth order, and evaluating them at zero, we have for the Taylor series of tangent:

$$ \tan(x) = x + \frac{2}{3!}x^3 + \frac{16}{5!}x^5 \cdots $$

The domain of tangent is $(-\pi/2, \pi/2)$, which repeats every $\pi$ radians in both directions.

Inversion

The description of the inverse of Sine is with the understanding that an interval of $2\pi$ radians defines a domain of every period of the cyclical function.