Trigonometric Functions

Angles, oscillations, and rotations all involve the same pair of functions — sine and cosine. At the basis level these are most cleanly defined as specific power series, in direct parallel to how $\exp$ was defined in Exponential Functions. This approach avoids any appeal to geometry and gives you precise formulas for derivatives, identities, and even the definition of $\pi$ — all from first principles.

Power series definitions

Recall from Exponential Functions that $\exp(x) = \sum_{k=0}^{\infty} x^k/k!$ converges absolutely for every $x \in \mathbb{R}$. Define two new functions by splitting this series into its even-indexed and odd-indexed terms, each with alternating signs:

$$\cos x \;\coloneqq\; \sum_{k=0}^{\infty} \frac{(-1)^k\,x^{2k}}{(2k)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \tag{1}$$ $$\sin x \;\coloneqq\; \sum_{k=0}^{\infty} \frac{(-1)^k\,x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \tag{2}$$

Both series converge absolutely for all $x \in \mathbb{R}$: the ratio test gives a ratio $|x|^2\!/\bigl((2k+1)(2k+2)\bigr) \to 0$ as $k \to \infty$, so convergence is absolute and uniform on every bounded interval. Uniform convergence justifies differentiating the series term-by-term.

At $x = 0$: the constant term of the cosine series is $1$ and all other terms vanish, giving $\cos(0) = 1$. The sine series starts with $x$, so $\sin(0) = 0$.

Motivation: Euler’s formula

Why these particular series? Substitute $ix$ (where $i^2 = -1$) formally into the exponential series:

$$\exp(ix) = \sum_{k=0}^{\infty} \frac{(ix)^k}{k!}.$$

Separating even and odd powers, and using $i^{2m} = (-1)^m$ and $i^{2m+1} = i(-1)^m$, the real part is exactly the cosine series $(1)$ and the imaginary part is $i$ times the sine series $(2)$. This yields Euler’s formula:

$$\exp(ix) = \cos x + i\sin x. \tag{3}$$

At the basis level, treat $(3)$ as a motivating insight rather than a fully rigorous statement (a complete treatment requires extending $\exp$ to the complex numbers). The formula makes the connection between exponentials and trigonometry vivid and will let you derive the addition formulas below in a single line.

The special case $x = \pi$ gives Euler’s identity $\exp(i\pi) + 1 = 0$, once $\pi$ is defined — which you will do shortly.

Derivatives

Differentiate the series $(1)$ and $(2)$ term-by-term:

$$\frac{d}{dx}\cos x = \sum_{k=1}^{\infty} \frac{(-1)^k \cdot 2k \cdot x^{2k-1}}{(2k)!} = -\sum_{j=0}^{\infty} \frac{(-1)^j\,x^{2j+1}}{(2j+1)!} = -\sin x,$$ $$\frac{d}{dx}\sin x = \sum_{k=0}^{\infty} \frac{(-1)^k(2k+1)x^{2k}}{(2k+1)!} = \sum_{k=0}^{\infty} \frac{(-1)^k\,x^{2k}}{(2k)!} = \cos x.$$

In summary:

$$(\cos x)' = -\sin x, \qquad (\sin x)' = \cos x. \tag{4}$$

Applying $(4)$ twice: $(\cos x)'' = -\cos x$ and $(\sin x)'' = -\sin x$. Both functions therefore satisfy the simple harmonic oscillator equation $y'' + y = 0$.

The Pythagorean identity

Theorem. For all $x \in \mathbb{R}$:

$$\cos^2 x + \sin^2 x = 1. \tag{5}$$

Proof. Let $f(x) \coloneqq \cos^2 x + \sin^2 x$. Differentiate using $(4)$:

$$f'(x) = 2\cos x \cdot (-\sin x) + 2\sin x \cdot \cos x = 0.$$

So $f$ is constant on $\mathbb{R}$. Evaluating at $x = 0$:

$$f(0) = \cos^2(0) + \sin^2(0) = 1^2 + 0^2 = 1.$$

Therefore $f(x) = 1$ for all $x$. $\square$

The Pythagorean identity $(5)$ is the source of almost every trigonometric manipulation you will encounter. One immediate consequence: $|\cos x| \leq 1$ and $|\sin x| \leq 1$ for all $x$.

Addition formulas

Theorem. For all $x, y \in \mathbb{R}$:

$$\cos(x + y) = \cos x \cos y - \sin x \sin y, \tag{6}$$ $$\sin(x + y) = \sin x \cos y + \cos x \sin y. \tag{7}$$

Proof using Euler’s formula. Multiply the two complex exponentials:

$$\exp(i(x + y)) = \exp(ix)\exp(iy) = (\cos x + i\sin x)(\cos y + i\sin y).$$

Expanding the right side:

$$= (\cos x\cos y - \sin x\sin y) + i(\sin x\cos y + \cos x\sin y).$$

The left side equals $\cos(x+y) + i\sin(x+y)$ by $(3)$. Equating real and imaginary parts gives $(6)$ and $(7)$. $\square$

Setting $y = x$ in $(6)$ gives the double-angle formula $\cos(2x) = \cos^2 x - \sin^2 x$. Combined with the Pythagorean identity $(5)$, this also gives $\cos(2x) = 2\cos^2 x - 1 = 1 - 2\sin^2 x$.

Defining $\pi$

The cosine series $(1)$ gives $\cos(0) = 1 > 0$. You can show that $\cos(2) < 0$ by examining the alternating partial sums: grouping consecutive pairs shows every partial sum from the second term onward is negative when $x = 2$, and a careful estimate gives $\cos(2) < -\tfrac{1}{3}$.

Since $\cos$ is continuous (its series converges uniformly), the Intermediate Value Theorem guarantees at least one zero of $\cos$ in the interval $(0, 2)$.

Definition. $\boldsymbol{\pi}$ is defined as twice the smallest positive zero of $\cos$:

$$\pi \;\coloneqq\; 2\cdot\min\{\,x > 0 \mid \cos x = 0\,\}. \tag{8}$$

From this definition, $\cos(\pi/2) = 0$. Applying the Pythagorean identity $(5)$ at $x = \pi/2$ forces $\sin^2(\pi/2) = 1$. Because $\sin$ is positive on $(0,\,\pi/2)$ — its derivative there equals $\cos > 0$, and $\sin(0) = 0$ — you get $\sin(\pi/2) = 1$.

Using the addition formulas $(6)$ and $(7)$ with $y = \pi/2$:

$$\cos\!\left(x + \frac{\pi}{2}\right) = -\sin x, \qquad \sin\!\left(x + \frac{\pi}{2}\right) = \cos x.$$

Applying this shift twice gives $\cos(x + \pi) = -\cos x$ and $\sin(x + \pi) = -\sin x$, and one more application gives periodicity.

Periodicity

Theorem. For all $x \in \mathbb{R}$:

$$\cos(x + 2\pi) = \cos x, \qquad \sin(x + 2\pi) = \sin x. \tag{9}$$

Proof. Apply $\cos(x + \pi) = -\cos x$ twice:

$$\cos(x + 2\pi) = \cos\bigl((x+\pi)+\pi\bigr) = -\cos(x+\pi) = -(-\cos x) = \cos x,$$

and similarly for $\sin$. $\square$

In fact $2\pi$ is the minimal period of both functions, though proving this requires confirming that $\cos$ has no positive zeros smaller than $\pi/2$.

Values at special angles

The definitions and identities you have developed so far pin down the values at the standard angles:

$x$	$\cos x$	$\sin x$
$0$	$1$	$0$
$\pi/6$	$\sqrt{3}/2$	$1/2$
$\pi/4$	$1/\sqrt{2}$	$1/\sqrt{2}$
$\pi/3$	$1/2$	$\sqrt{3}/2$
$\pi/2$	$0$	$1$
$\pi$	$-1$	$0$

For $\pi/4$: the addition formula with $x = y = \pi/4$ and $\cos(\pi/2) = 0$ gives $\cos^2(\pi/4) = \sin^2(\pi/4)$; combined with $(5)$, both equal $1/\sqrt{2}$.

For $\pi/3$: the double-angle formula gives $\cos(2\pi/3) = 2\cos^2(\pi/3) - 1$. Because $\cos(2\pi/3) = \cos(\pi - \pi/3) = -\cos(\pi/3)$ (a consequence of $\cos(x+\pi) = -\cos x$), solving $-\cos(\pi/3) = 2\cos^2(\pi/3) - 1$ yields $\cos(\pi/3) = 1/2$ and then $\sin(\pi/3) = \sqrt{3}/2$ from $(5)$.

Other trigonometric functions

Four more functions are defined as ratios and reciprocals of sine and cosine.

Tangent:

$$\tan x \;\coloneqq\; \frac{\sin x}{\cos x}, \qquad \cos x \neq 0.$$

The domain of $\tan$ is $\mathbb{R} \setminus \{\pi/2 + k\pi \mid k \in \mathbb{Z}\}$ and its minimal period is $\pi$.

Cotangent:

$$\cot x \;\coloneqq\; \frac{\cos x}{\sin x}, \qquad \sin x \neq 0.$$

Secant and cosecant are the reciprocals:

$$\sec x \;\coloneqq\; \frac{1}{\cos x}, \qquad \csc x \;\coloneqq\; \frac{1}{\sin x}.$$

Derivatives of $\tan$ and $\cot$

Apply the quotient rule, then use the Pythagorean identity $(5)$:

$$(\tan x)' = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x. \tag{10}$$ $$(\cot x)' = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = -\frac{\cos^2 x + \sin^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x. \tag{11}$$

The inverse trigonometric functions — $\arcsin$, $\arccos$, $\arctan$ — are defined and studied in Inverse Trigonometric Functions.

Summary

• Cosine and sine are defined by the absolutely convergent power series $(1)$ and $(2)$; at $x = 0$, $\cos(0) = 1$ and $\sin(0) = 0$.
• Euler’s formula $\exp(ix) = \cos x + i\sin x$ arises by separating real and imaginary parts of the exponential series after substituting $ix$.
• Derivatives: $(\cos x)' = -\sin x$ and $(\sin x)' = \cos x$; both functions satisfy $y'' + y = 0$.
• Pythagorean identity: $\cos^2 x + \sin^2 x = 1$, proved by showing the derivative of the left side is $0$ and evaluating at $x = 0$.
• Addition formulas $(6)$ and $(7)$ are derived by expanding $\exp(ix)\exp(iy) = \exp(i(x+y))$.
• $\pi$ is defined as twice the smallest positive zero of $\cos$; its existence follows from the Intermediate Value Theorem applied to the continuous function $\cos$.
• Periodicity: $\cos$ and $\sin$ each have period $2\pi$.
• Key values: $\cos(\pi/2) = 0$, $\sin(\pi/2) = 1$, $\cos(\pi) = -1$, $\sin(\pi) = 0$, $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$, $\cos(\pi/3) = 1/2$, $\sin(\pi/3) = \sqrt{3}/2$.
• $\tan x = \sin x/\cos x$ with $(\tan x)' = \sec^2 x$; $\cot x = \cos x/\sin x$ with $(\cot x)' = -\csc^2 x$.