Fundamental Trigonometric Properties#

Theorem: Trigonometric Functions of Standard Angles

Here are some common values for sine, cosine, tangent and cotangent:

	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\frac{2\pi}{3}$	$\frac{5\pi}{6}$	$\pi$
$\sin \theta$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$0$
$\cos \theta$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$	$-\frac{1}{2}$	$-\frac{\sqrt{3}}{2}$	$-1$
$\tan \theta$	$0$	$\frac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$		$-\sqrt{3}$	$-\frac{\sqrt{3}}{3}$	$0$
$\cot \theta$		$\sqrt{3}$	$1$	$\frac{\sqrt{3}}{3}$	$0$	$-\frac{\sqrt{3}}{3}$	$-\sqrt{3}$

Proof

TODO

Theorem: Fundamental Trigonometric Identities

Sine, cosine, tangent and cotangent obey the following identities:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

\[ \tan \theta \cot \theta = 1 \]

Proof

Proof of (1):

We define $f(\theta) = \sin^2 \theta + \cos^2 \theta$ and then differentiate $f$:

\[ \begin{aligned} f'(\theta) &= (\sin^2 \theta + \cos^2 \theta)' \\ &= (\sin^2 \theta)' + (\cos^2 \theta)' \\ &= 2\sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0 \end{aligned} \]

Since the derivative of $f$ is always zero, we know that $f$ is a constant function, i.e. $f(\theta) = C$ for all $\theta$. To find $C$, we just need to evaluate $f$ at any convenient point, for example $\theta = 0$.

\[ f(0) = \sin^2(0) + \cos^2(0) = 0 + 1 = 1 \]

Proof of (2):

This follows directly from the definitions of tangent and cotangent.

Theorem: Universal Trigonometric Substitution

The sine, cosine, tangent and cotangent of $\theta$ can all be expressed in terms of $\tan \frac{\theta}{2}$:

\[ \sin \theta = \frac{2\tan \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}} \]

\[ \cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} \]

\[ \tan \theta = \frac{2\tan \frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}} \]

\[ \cot \theta = \frac{1-\tan^2\frac{\theta}{2}}{2\tan \frac{\theta}{2}} \]

Proof

Proof of (1):

TODO

Angle Sums#

Theorem: Trigonometric Identities for Angle Sums

Sine, cosine, tangent and cotangent have the following properties:

\[ \begin{aligned} \sin(\theta \pm \varphi) &= \sin\theta\cos\varphi \pm \cos\theta \sin \varphi \\ \\ \cos (\theta \pm \varphi) &= \cos \theta \cos \varphi \mp \sin \theta \sin \varphi \\ \\ \tan (\theta \pm \varphi) &= \frac{\tan \theta \pm \tan \varphi}{1 \mp \tan \theta \tan \varphi} \\ \\ \cot(\theta \pm \varphi) &= \frac{\cot \theta \cot \varphi \mp 1}{\cot \varphi \pm \cot \theta} \end{aligned} \]

Proof

Proof of (1):

From the definition of sine:

\[ \sin(\theta + \varphi) = \sum_{n = 0}^\infty (-1)^n \frac{(\theta+\varphi)^{2n+1}}{(2n+1)!} \]

Using the binomial theorem, we can rewrite $(\theta+\varphi)^{2n+1}$ as

\[ (\theta+\varphi)^{2n+1} = \sum_{k = 0}^{2n+1} \binom{2n+1}{k} \theta^k \varphi^{2n+1 - k} \]

Therefore,

\[ \sin(\theta + \varphi) = \sum_{n = 0}^\infty \left[\frac{(-1)^n}{(2n+1)!} \sum_{k = 0}^{2n+1} \binom{2n+1}{k} \theta^k \varphi^{2n+1 - k}\right] \]

We notice that when $k$ is odd, $(2n+1 - k)$ is even and, when $k$ is even, $(2n + 1 - k)$ is odd. We thus split the inner sum (the one over $k$) into two sums, where the first contains only the odd powers of $\theta$ and the second contains only the even powers of $\theta$.

\[ \sum_{k = 0}^{2n+1} \binom{2n+1}{k} \theta^k \varphi^{2n+1 - k} = \sum_{j = 0}^n \binom{2n+1}{2j+1} \theta^{2j + 1} \varphi^{2n - 2j} + \sum_{j = 0}^n \binom{2n+1}{2j} \theta^{2j} \varphi^{2n + 1 - 2j} \]

We can substitute this into the formula for $\sin(\theta + \varphi)$:

\[ \sin(\theta + \varphi) = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)!}\left[ \sum_{j = 0}^n \binom{2n+1}{2j+1} \theta^{2j + 1} \varphi^{2n - 2j} + \sum_{j = 0}^n \binom{2n+1}{2j} \theta^{2j} \varphi^{2n + 1 - 2j}\right] \\ \]

Since the power series for sine is absolutely convergent everywhere, we can distribute the outer sum:

\[ \sin(\theta + \varphi) = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)!} \left[\sum_{j = 0}^n \binom{2n+1}{2j+1} \theta^{2j + 1} \varphi^{2n - 2j} \right] + \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)!} \left[\sum_{j = 0}^n \binom{2n+1}{2j} \theta^{2j} \varphi^{2n + 1 - 2j}\right] \]

We examine each part separately.

\[ \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)!} \left[\sum_{j = 0}^n \binom{2n+1}{2j+1} \theta^{2j + 1} \varphi^{2n - 2j} \right] = \sum_{n = 0}^\infty \sum_{j = 0}^n \frac{(-1)^n}{(2n+1)!} \binom{2n+1}{2j+1} \theta^{2j + 1} \varphi^{2n - 2j} \]

Expand the binomial coefficient:

\[ \sum_{n = 0}^\infty \sum_{j = 0}^n \frac{(-1)^n}{(2n+1)!} \binom{2n+1}{2j+1} \theta^{2j + 1} \varphi^{2n - 2j} = \sum_{n = 0}^\infty \sum_{j = 0}^n \frac{(-1)^n}{(2n+1)!} \frac{(2n+1)!}{(2j+1)!(2n - 2j)!} \theta^{2j + 1} \varphi^{2n - 2j} \]

Cancel the $(2n+1)!$ factors:

\[ \sum_{n = 0}^\infty \sum_{j = 0}^n \frac{(-1)^n}{(2n+1)!} \frac{(2n+1)!}{(2j+1)!(2n - 2j)!} \theta^{2j + 1} \varphi^{2n - 2j} = \sum_{n = 0}^\infty \sum_{j = 0}^n \frac{(-1)^n}{(2j+1)!(2n - 2j)!} \theta^{2j + 1} \varphi^{2n - 2j} \]

Let $m = n - j$. We can thus rewrite $(-1)^n$ as $(-1)^{j+m} = (-1)^j (-1)^m$ and obtain

\[ \begin{aligned} \sum_{n = 0}^\infty \sum_{j = 0}^n \frac{(-1)^n}{(2j+1)!(2n - 2j)!} \theta^{2j + 1} \varphi^{2n - 2j} &= \sum_{j = 0}^{\infty}\sum_{m = 0}^{\infty} (-1)^j (-1)^m \frac{\theta^{2j+1}}{(2j+1)!}\frac{\varphi^{2m}}{(2m)!} \\ &= \left(\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!}\theta^{2j + 1}\right) \left(\sum_{m = 0}^{\infty} \frac{(-1)^m}{(2m)!}\varphi^{2m}\right) \end{aligned} \]

The first parentheses contain the definition of $\sin \theta$ and the second parentheses contain the definition of $\cos \varphi$. Therefore,

\[ \sin(\theta + \varphi) = \sin \theta \cos \varphi + \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)!} \left[\sum_{j = 0}^n \binom{2n+1}{2j} \theta^{2j} \varphi^{2n + 1 - 2j}\right]. \]

The proof that the second sum is equal to $\cos \theta \sin \varphi$ is analogous.

To prove that $\sin(\theta - \varphi) = \sin \theta \cos \varphi - \cos \theta \sin \varphi$, we use the parity of sine and cosine:

\[ \sin(\theta - \varphi) = \sin(\theta + (-\varphi)) = \sin \theta \cos (-\varphi) + \cos \theta \sin (-\varphi) = \sin \theta \cos \varphi - \cos \theta \sin \varphi \]

Proof of (2):

We use the fact that $\cos(\theta \pm \varphi) = \sin\left(\frac{\pi}{2} - (\theta \pm \varphi)\right)$:

\[ \cos(\theta + \varphi) = \sin\left(\frac{\pi}{2} - (\theta + \varphi)\right) = \sin\left(\left(\frac{\pi}{2} - \theta\right) - \varphi\right) \]

Apply the formula for $\sin (x - y)$:

\[ \sin\left(\left(\frac{\pi}{2} - \theta\right) - \varphi\right) = \sin\left(\frac{\pi}{2} - \theta\right)\cos{\varphi} - \cos\left(\frac{\pi}{2} - \theta\right) \]

Again, we use the fact that $\sin\left(\frac{\pi}{2} - x\right) = \cos(x)$ and $\sin\left(\frac{\pi}{2} - x\right) = \cos(x)$:

\[ \sin\left(\frac{\pi}{2} - \theta\right)\cos{\varphi} - \cos\left(\frac{\pi}{2} - \theta\right) = \cos \theta \cos \varphi - \sin \theta \sin \varphi \]

To prove that $\cos(\theta - \varphi) = \cos \theta \cos \varphi + \sin \theta \sin \varphi$, we use the parity of sine and cosine:

\[ \cos(\theta - \varphi) = \cos(\theta + (-\varphi)) = \cos \theta \cos (-\varphi) - \sin \theta \sin (-\varphi) = \cos \theta \cos \varphi + \sin \theta \sin \varphi \]

Proof of (3):

We use the definition of tangent:

\[ \tan(\theta \pm \varphi) = \frac{\sin (\theta \pm \varphi)}{\cos(\theta \pm \varphi)} \]

Use the formulas for $\sin (x \pm y)$ and $\cos (x \pm y)$:

\[ \frac{\sin (\theta \pm \varphi)}{\cos(\theta \pm \varphi)} = \frac{\sin \theta \cos \varphi + \cos \theta \sin \varphi}{\cos \theta \cos \varphi \mp \sin \theta \sin \varphi} \]

We divide the numerator and denominator by $\cos \theta \cos \varphi$:

\[ \frac{\sin \theta \cos \varphi \pm \cos \theta \sin \varphi}{\cos \theta \cos \varphi \mp \sin \theta \sin \varphi} = \frac{\frac{\sin\theta \cos\varphi}{\cos\theta \cos\varphi} \pm \frac{\cos\theta \sin\varphi}{\cos\theta \cos\varphi}}{\frac{\cos\theta \cos\varphi}{\cos\theta \cos\varphi} \mp \frac{\sin\theta \sin\varphi}{\cos\theta \cos\varphi}} \]

Cancel like terms:

\[ \frac{\frac{\sin\theta \cos\varphi}{\cos\theta \cos\varphi} \pm \frac{\cos\theta \sin\varphi}{\cos\theta \cos\varphi}}{\frac{\cos\theta \cos\varphi}{\cos\theta \cos\varphi} \mp \frac{\sin\theta \sin\varphi}{\cos\theta \cos\varphi}} = \frac{\frac{\sin\theta}{\cos\theta} \pm \frac{\sin\varphi}{\cos\varphi}}{1 \mp \left(\frac{\sin\theta}{\cos\theta}\right)\left(\frac{\sin\varphi}{\cos\varphi}\right)} \]

Use the definition of tangent again:

\[ \frac{\frac{\sin\theta}{\cos\theta} \pm \frac{\sin\varphi}{\cos\varphi}}{1 \mp \left(\frac{\sin\theta}{\cos\theta}\right)\left(\frac{\sin\varphi}{\cos\varphi}\right)} = \frac{\tan \theta \pm \tan \varphi}{1 \mp \tan \theta \tan \varphi} \]

Proof of (4):

We use the definition of cotangent:

\[ \cot (\theta \pm \varphi) = \frac{\cos(\theta \pm \varphi)}{\sin(\theta \pm \varphi)} \]

Apply the formulas for $\cos(\theta \pm \varphi)$ and $\sin(\theta \pm \varphi)$:

\[ \frac{\cos (\theta \pm \varphi)}{\sin(\theta \pm \varphi)} = \frac{\cos \theta \cos \varphi \mp \sin \theta \sin \varphi}{\sin \theta \cos \varphi \pm \cos \theta \sin \varphi} \]

We divide the numerator and denominator by $\sin \theta \sin \varphi$:

\[ \frac{\cos \theta \cos \varphi \mp \sin \theta \sin \varphi}{\sin \theta \cos \varphi \pm \cos \theta \sin \varphi} = \frac{\frac{\cos \theta \cos \varphi}{\sin \theta \sin \varphi} \mp \frac{\sin \theta \sin \varphi}{\sin \theta \sin \varphi}}{\frac{\sin \theta \cos \varphi}{\sin \theta \sin \varphi} \pm \frac{\cos \theta \sin \varphi}{\sin \theta \sin \varphi}} \]

Cancel like terms:

\[ \frac{\frac{\cos \theta \cos \varphi}{\sin \theta \sin \varphi} \mp \frac{\sin \theta \sin \varphi}{\sin \theta \sin \varphi}}{\frac{\sin \theta \cos \varphi}{\sin \theta \sin \varphi} \pm \frac{\cos \theta \sin \varphi}{\sin \theta \sin \varphi}} = \frac{\frac{\cos \theta}{\sin \theta} \frac{\cos \varphi}{\sin \varphi} \mp 1}{\frac{\cos \varphi}{\sin \varphi} \pm \frac{\cos \theta}{\sin \theta}} \]

Use the definition of cotangent again:

\[ \frac{\frac{\cos \theta}{\sin \theta} \frac{\cos \varphi}{\sin \varphi} \mp 1}{\frac{\cos \varphi}{\sin \varphi} \pm \frac{\cos \theta}{\sin \theta}} = \frac{\cot \theta \cot \varphi \mp 1}{\cot \varphi \pm \cot \theta} \]

Angle Products#

Theorem: Chebyshev's Formulas

The sine, cosine and tangent obey Chebyshev's formulas for every $r \in \mathbb{R}$:

\[ \begin{aligned}\sin(r\varphi) &= 2\cos \varphi \sin ((r-1)\varphi) - \sin ((r-2)\varphi) \\ \\ \cos (r \varphi) &= 2\cos \varphi \cos ((r-1)\varphi) - \cos((r-2)\varphi) \\ \\ \tan(r\varphi) &= \frac{\tan((r-1)\varphi) + \tan \varphi}{1 - \tan \varphi\tan ((r-1)\varphi)} \end{aligned} \]

Proof

TODO

Theorem: Double-Angle Formulas

The sine, cosine, tangent and cotangent of $2\theta$ can be expressed as

\[ \begin{aligned} \sin(2\theta) &= 2\sin \theta \cos \theta \\ \\ \cos(2\theta) &= 2\cos^2\theta + 1 = \cos^2 \theta - \sin^2 \theta = 1-2\sin^2 \theta \\ \\ \tan (2\theta) &= \frac{2\tan \theta}{1-\tan^2 \theta} \end{aligned} \]

Proof

Proof of (1):

We use the summation formula:

\[ \sin(2\theta) = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2 \sin \theta \cos \theta \]

Proof of (2):

TODO

Proof of (3):

We use the summation formula:

\[ \tan(2 \theta) = \tan(\theta + \theta) = \frac{\tan \theta + \tan \theta}{1 - \tan \theta \tan \theta} = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]

Theorem: Half-Angle Formulae

The sine, cosine, tangent and cotangent obey the following properties:

\[ \begin{aligned}\sin \frac{\theta}{2} &= \operatorname{sgn}\left(\sin \frac{\theta}{2}\right) \sqrt{\frac{1-\cos\theta}{2}} \\ \\ \cos \frac{\theta}{2} &= \operatorname{sgn}\left(\cos \frac{\theta}{2}\right) \sqrt{\frac{1+\cos\theta}{2}} \\ \\ \tan\frac{\theta}{2} &= \frac{\sin \theta}{1+\cos \theta} = \frac{1 - \cos \theta}{\sin \theta} = \operatorname{sgn}(\sin \theta) \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \\ \\ \cot \frac{\theta}{2} &= \frac{1+\cos \theta}{\sin \theta} = \frac{\sin \theta}{1 - \cos \theta} = \pm \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} \end{aligned} \]

Proof

TODO

Argument Offsets#

Theorem: Trigonometric Identities for Argument Offsets

The sine, cosine, tangent and cotangent have the following properties:

	$\frac{\pi}{2}+\theta$	$\frac{\pi}{2}-\theta$	$\pi+\theta$	$\pi-\theta$	$\frac{3\pi}{2} + \theta$	$\frac{3\pi}{2} - \theta$
$\sin$	$\cos \theta$	$\cos \theta$	$-\sin \theta$	$\sin \theta$	$-\cos \theta$	$-\cos \theta$
$\cos$	$-\sin \theta$	$\sin \theta$	$-\cos \theta$	$-\cos \theta$	$\sin \theta$	$-\sin \theta$
$\tan$	$-\cot \theta$	$\cot \theta$	$\tan \theta$	$-\tan \theta$	$-\cot \theta$	$\cot \theta$
$\cot$	$-\tan \theta$	$\tan \theta$	$\cot \theta$	$-\cot \theta$	$-\tan \theta$	$\tan \theta$

Proof

TODO

Function Sums#

Theorem: Trigonometric Identities for Function Sums

sine, cosine, tangent and cotangent have the following properties:

\[ \begin{aligned} \sin \theta \pm \sin \varphi &= 2\sin \frac{\theta \pm \varphi}{2} \cos \frac{\theta \mp \varphi}{2} \\ \\ \cos \theta + \cos \varphi &= 2 \cos \frac{\theta + \varphi}{2} \cos \frac{\theta - \varphi}{2} \\ \\ \cos \theta - \cos \varphi &= -2 \sin \frac{\theta + \varphi}{2} \sin \frac{\theta - \varphi}{2} \\ \\ \tan \theta \pm \tan \varphi &= \frac{\sin(\theta \pm \varphi)}{\cos \theta \cos \varphi} \\ \\ \cot \theta \pm \cot \varphi &= \frac{\sin(\varphi \pm \theta)}{\sin \theta \sin \varphi} \end{aligned} \]

Proof

Proof of (1):

Proof of (2):

Proof of (3):

Proof of (4):

We use the definition of tangent and the formula for $\sin (x \pm y)$:

\[ \begin{aligned} \tan \theta \pm \tan \varphi &= \frac{\sin \theta}{\cos \theta} \pm \frac{\sin \varphi}{\cos \varphi} \\ &= \frac{\sin \theta \cos \varphi \pm \sin \varphi \cos \theta}{\cos \theta \cos \varphi} \\ &= \frac{\sin (\theta \pm \varphi)}{\cos \theta \cos \varphi} \end{aligned} \]

Proof of (5):

TODO

Function Products#

Theorem: Trigonometric Identities for Function Products

sine, cosine, tangent and cotangent have the following properties:

\[ \begin{aligned} \cos \theta \cos \varphi &= \frac{1}{2}(\cos(\theta - \varphi) + \cos(\theta + \varphi)) \\ \\ \sin \theta \sin \varphi &= \frac{1}{2}(\cos(\theta - \varphi) - \cos(\theta + \varphi)) \\ \\ \sin \theta \cos \varphi &= \frac{1}{2}(\sin(\theta + \varphi) + \sin(\theta - \varphi)) \\ \\ \tan \theta \tan \varphi &= \frac{\cos(\theta - \varphi) - \cos (\theta + \varphi)}{\cos(\theta - \varphi) + \cos (\theta + \varphi)} \\ \\ \tan \theta \cot \varphi &= \frac{\sin(\theta + \varphi) + \sin(\theta - \varphi)}{\sin(\theta + \varphi) - \sin(\theta - \varphi)}\end{aligned} \]

Proof

Proof of (1):

We use the formula for $\cos(x \pm y)$:

\[ \begin{aligned} \frac{1}{2}(\cos(\theta - \varphi) + \cos(\theta + \varphi)) &= \frac{1}{2}(\cos \theta \cos \varphi + \sin \theta \sin \varphi + \cos \theta \cos \varphi - \sin \theta \sin \varphi) \\ &= \frac{1}{2} \cdot 2 \cos \theta \cos \varphi \\ &= \cos \theta \cos \varphi \end{aligned} \]

Proof of (2):

We use the formula for $\cos(x \pm y)$:

\[ \begin{aligned} \frac{1}{2}(\cos(\theta - \varphi) - \cos(\theta + \varphi)) &= \frac{1}{2}(\cos \theta \cos \varphi + \sin \theta \sin \varphi - ( \cos \theta \cos \varphi - \sin \theta \sin \varphi)) \\ &= \frac{1}{2}(\cos \theta \cos \varphi + \sin \theta \sin \varphi - \cos \theta \cos \varphi + \sin \theta \sin \varphi) \\ &= \frac{1}{2} \cdot 2 \sin \theta \sin \varphi \\ &= \sin \theta \sin \varphi \end{aligned} \]

Proof of (3):

We use the formula for $\sin(x \pm y)$:

\[ \frac{1}{2}(\sin(\theta + \varphi) + \sin(\theta - \varphi)) = \frac{1}{2}(\sin \theta \cos \varphi + \cos \theta \sin \varphi + \sin \theta \cos \varphi - \cos \theta \sin \varphi) = \frac{1}{2}\cdot 2 \sin \theta \cos \varphi = \sin \theta \cos \varphi \]

Proof of (4):

We use the definition of tangent and the formulas for $\sin \theta \sin \varphi$ and $\cos \theta \cos \varphi$:

\[ \begin{aligned} \tan \theta \tan \varphi &= \frac{\sin \theta}{\cos \theta}\frac{\sin \varphi}{\cos \varphi} \\ &= \frac{\sin \theta \sin \varphi}{\cos \theta \cos \varphi} \\ &= \frac{\frac{1}{2}(\cos(\theta - \varphi) - \cos(\theta + \varphi))}{\frac{1}{2}(\cos(\theta - \varphi) + \cos(\theta + \varphi))} \\ &= \frac{\cos(\theta - \varphi) - \cos (\theta + \varphi)}{\cos(\theta - \varphi) + \cos (\theta + \varphi)} \end{aligned} \]

Proof of (5):

We use the definition of tangent and cotangent and the formula for $\sin \theta \sin \varphi$ and $\cos \theta \cos \varphi$:

\[ \begin{aligned} \tan \theta \cot \varphi &= \frac{\sin \theta}{\cos \theta}\frac{\cos \varphi}{\sin \varphi} \\ &= \frac{\sin \theta \cos \varphi}{\cos \theta \sin \varphi} \\ &= \frac{\frac{1}{2}(\sin(\theta + \varphi) + \sin(\theta - \varphi))}{\frac{1}{2}(\sin(\varphi + \theta) + \sin(\varphi - \theta))} \\ &= \frac{\sin(\theta + \varphi) + \sin(\theta - \varphi)}{\sin(\varphi + \theta) + \sin(\varphi - \theta)} \\ &= \frac{\sin(\theta + \varphi) + \sin(\theta - \varphi)}{\sin(\theta + \varphi) - \sin(\theta - \varphi)} \end{aligned} \]

Compositions#

Theorem: Functions of Arcfunctions

The compositions of inverse real trigonometric functions inside real trigonometric functions have the following properties:

$$
\begin{aligned}

&\sin(\arctan x) = \frac{x}{\sqrt{1+x^2}} \

&\cos(\arctan (x)) = \frac{1}{\sqrt{1 + x^2}} \

&\cos(\arcsin(x)) = \sqrt{1 - x^2} \

&\cos\left(\arctan \left( \frac{1}{x} \right) \right) = \frac{|x|}{\sqrt{1 + x^2}}
\end{aligned}
$$

Proof

Proof of (1):

We substitute $y = \arctan x$ and so $x = \tan y$. Therefore, we have

\[ \sin(\arctan x) = \frac{x}{\sqrt{1+x^2}} \iff \sin y = \frac{\tan y}{\sqrt{1 + \tan^2 y}} \]

Now, we use the definition of $\tan y$:

\[ \begin{aligned} \frac{\tan y}{\sqrt{1 + \tan^2 y}} &= \frac{\frac{\sin y}{\cos y}}{\sqrt{1 + \frac{\sin^2 y}{\cos^2 y}}} \\ &= \frac{\sin y}{\cos y} \times \frac{1}{\sqrt{1 + \frac{\sin^2 y}{\cos^2 y}}} \\ &= \frac{\sin y}{\cos y} \times \frac{1}{\sqrt{ \frac{\cos^2 y + \sin^2 y}{\cos^2 y}}} \\ &= \frac{\sin y}{\cos y} \times \frac{1}{\sqrt{ \frac{1}{\cos^2 y}}} \\ &= \frac{\sin y}{\cos y} \times \sqrt{\cos^2 y} \end{aligned} \]

Since $y = \arctan x$ and the image of $\arctan$ is $\left(-\frac{\pi}{2}; \frac{\pi}{2}\right)$, we know that $y \in \left(-\frac{\pi}{2}; \frac{\pi}{2}\right)$. For $y = \left(-\frac{\pi}{2}; \frac{\pi}{2}\right)$, we know that $\cos y \gt 0$, so we can cancel the square root and the second power.

\[ \frac{\sin y}{\cos y} \times \sqrt{\cos^2 y} = \frac{\sin y}{\cos y} \times \cos y = \sin y \]

Proof of (3):

From $\sin^2 y + \cos^2 y = 1$, we get

\[ \sin^2 (\arcsin x) + \cos^2(\arcsin x) = 1 \]

for some $x = \sin (y)$. We thus get

\[ \begin{aligned} &x^2 + \cos^2(\arcsin x) = 1 \\ &\cos^2(\arcsin x) = 1 - x^2 \end{aligned} \]

The right-hand side is always non-negative for the domain of $\arcsin$, i.e. $x \in [-1; +1]$, so we can take the square root:

\[ \cos (\arcsin x) = \sqrt{1-x^2} \]

TODO

	\(0\)	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{\pi}{3}\)	\(\frac{\pi}{2}\)	\(\frac{2\pi}{3}\)	\(\frac{5\pi}{6}\)	\(\pi\)
\(\sin \theta\)	\(0\)	\(\frac{1}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{\sqrt{3}}{2}\)	\(1\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{1}{2}\)	\(0\)
\(\cos \theta\)	\(1\)	\(\frac{\sqrt{3}}{2}\)	\(\frac{\sqrt{2}}{2}\)	\(\frac{1}{2}\)	\(0\)	\(-\frac{1}{2}\)	\(-\frac{\sqrt{3}}{2}\)	\(-1\)
\(\tan \theta\)	\(0\)	\(\frac{\sqrt{3}}{3}\)	\(1\)	\(\sqrt{3}\)		\(-\sqrt{3}\)	\(-\frac{\sqrt{3}}{3}\)	\(0\)
\(\cot \theta\)		\(\sqrt{3}\)	\(1\)	\(\frac{\sqrt{3}}{3}\)	\(0\)	\(-\frac{\sqrt{3}}{3}\)	\(-\sqrt{3}\)

	\(\frac{\pi}{2}+\theta\)	\(\frac{\pi}{2}-\theta\)	\(\pi+\theta\)	\(\pi-\theta\)	\(\frac{3\pi}{2} + \theta\)	\(\frac{3\pi}{2} - \theta\)
\(\sin\)	\(\cos \theta\)	\(\cos \theta\)	\(-\sin \theta\)	\(\sin \theta\)	\(-\cos \theta\)	\(-\cos \theta\)
\(\cos\)	\(-\sin \theta\)	\(\sin \theta\)	\(-\cos \theta\)	\(-\cos \theta\)	\(\sin \theta\)	\(-\sin \theta\)
\(\tan\)	\(-\cot \theta\)	\(\cot \theta\)	\(\tan \theta\)	\(-\tan \theta\)	\(-\cot \theta\)	\(\cot \theta\)
\(\cot\)	\(-\tan \theta\)	\(\tan \theta\)	\(\cot \theta\)	\(-\cot \theta\)	\(-\tan \theta\)	\(\tan \theta\)