Trigonometric Properties

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 $f$:

$$\begin{array}{rl}{f}^{\prime }(\theta )& =({\sin }^2\theta +{\cos }^2\theta {)}^{\prime }\\ & =({\sin }^2\theta {)}^{\prime }+({\cos }^2\theta {)}^{\prime }\\ & =2\sin \theta \cos \theta -2\cos \theta \sin \theta =0\end{array}$$

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:

$$\begin{array}{rl}\sin (\theta \pm \phi )& =\sin \theta \cos \phi \pm \cos \theta \sin \phi \\ & \\ \cos (\theta \pm \phi )& =\cos \theta \cos \phi \mp \sin \theta \sin \phi \\ & \\ \tan (\theta \pm \phi )& =\frac{\tan \theta \pm \tan \phi }{1\mp \tan \theta \tan \phi }\\ & \\ \cot (\theta \pm \phi )& =\frac{\cot \theta \cot \phi \mp 1}{\cot \phi \pm \cot \theta }\end{array}$$

PROOF

Proof of (1):

From the definition of sine:

$$\sin (\theta +\phi )=\sum _{n=0}^{\infty }(-1{)}^n\frac{(\theta +\phi {)}^{2n+1}}{(2n+1)!}$$

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

$$(\theta +\phi {)}^{2n+1}=\sum _{k=0}^{2n+1}(\genfrac{}{}{0ex}{}{2n+1}{k}){\theta }^k{\phi }^{2n+1-k}$$

Therefore,

$$\sin (\theta +\phi )=\sum _{n=0}^{\infty }\left[\frac{(-1{)}^n}{(2n+1)!}\sum _{k=0}^{2n+1}\left(\genfrac{}{}{0ex}{}{2n+1}{k}\right){\theta }^k{\phi }^{2n+1-k}\right]$$

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 $\theta$.

$$\sum _{k=0}^{2n+1}\left(\genfrac{}{}{0ex}{}{2n+1}{k}\right){\theta }^k{\phi }^{2n+1-k}=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j+1}\right){\theta }^{2j+1}{\phi }^{2n-2j}+\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j}\right){\theta }^{2j}{\phi }^{2n+1-2j}$$

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

$$\sin (\theta +\phi )=\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)!}\left[\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j+1}\right){\theta }^{2j+1}{\phi }^{2n-2j}+\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j}\right){\theta }^{2j}{\phi }^{2n+1-2j}\right]$$

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

$$\sin (\theta +\phi )=\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)!}\left[\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j+1}\right){\theta }^{2j+1}{\phi }^{2n-2j}\right]+\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)!}\left[\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j}\right){\theta }^{2j}{\phi }^{2n+1-2j}\right]$$

We examine each part separately.

$$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)!}\left[\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j+1}\right){\theta }^{2j+1}{\phi }^{2n-2j}\right]=\sum _{n=0}^{\infty }\sum _{j=0}^n\frac{(-1{)}^n}{(2n+1)!}\left(\genfrac{}{}{0ex}{}{2n+1}{2j+1}\right){\theta }^{2j+1}{\phi }^{2n-2j}$$

Expand the binomial coefficient:

$$\sum _{n=0}^{\infty }\sum _{j=0}^n\frac{(-1{)}^n}{(2n+1)!}\left(\genfrac{}{}{0ex}{}{2n+1}{2j+1}\right){\theta }^{2j+1}{\phi }^{2n-2j}=\sum _{n=0}^{\infty }\sum _{j=0}^n\frac{(-1{)}^n}{(2n+1)!}\frac{(2n+1)!}{(2j+1)!(2n-2j)!}{\theta }^{2j+1}{\phi }^{2n-2j}$$

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

$$\sum _{n=0}^{\infty }\sum _{j=0}^n\frac{(-1{)}^n}{(2n+1)!}\frac{(2n+1)!}{(2j+1)!(2n-2j)!}{\theta }^{2j+1}{\phi }^{2n-2j}=\sum _{n=0}^{\infty }\sum _{j=0}^n\frac{(-1{)}^n}{(2j+1)!(2n-2j)!}{\theta }^{2j+1}{\phi }^{2n-2j}$$

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

$$\begin{array}{rl}\sum _{n=0}^{\infty }\sum _{j=0}^n\frac{(-1{)}^n}{(2j+1)!(2n-2j)!}{\theta }^{2j+1}{\phi }^{2n-2j}& =\sum _{j=0}^{\infty }\sum _{m=0}^{\infty }(-1{)}^j(-1{)}^m\frac{{\theta }^{2j+1}}{(2j+1)!}\frac{{\phi }^{2m}}{(2m)!}\\ & =\left(\sum _{j=0}^{\infty }\frac{(-1{)}^j}{(2j+1)!}{\theta }^{2j+1}\right)\left(\sum _{m=0}^{\infty }\frac{(-1{)}^m}{(2m)!}{\phi }^{2m}\right)\end{array}$$

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

$$\sin (\theta +\phi )=\sin \theta \cos \phi +\sum _{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)!}\left[\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2j}\right){\theta }^{2j}{\phi }^{2n+1-2j}\right].$$

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

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

$$\sin (\theta -\phi )=\sin (\theta +(-\phi ))=\sin \theta \cos (-\phi )+\cos \theta \sin (-\phi )=\sin \theta \cos \phi -\cos \theta \sin \phi$$

Proof of (2):

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

$$\cos (\theta +\phi )=\sin \left(\frac{\pi }{2}-(\theta +\phi )\right)=\sin \left(\left(\frac{\pi }{2}-\theta \right)-\phi \right)$$

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

$$\sin \left(\left(\frac{\pi }{2}-\theta \right)-\phi \right)=\sin \left(\frac{\pi }{2}-\theta \right)\cos \phi -\cos \left(\frac{\pi }{2}-\theta \right)$$

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

$$\sin \left(\frac{\pi }{2}-\theta \right)\cos \phi -\cos \left(\frac{\pi }{2}-\theta \right)=\cos \theta \cos \phi -\sin \theta \sin \phi$$

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

$$\cos (\theta -\phi )=\cos (\theta +(-\phi ))=\cos \theta \cos (-\phi )-\sin \theta \sin (-\phi )=\cos \theta \cos \phi +\sin \theta \sin \phi$$

Proof of (3):

We use the definition of tangent:

$$\tan (\theta \pm \phi )=\frac{\sin (\theta \pm \phi )}{\cos (\theta \pm \phi )}$$

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

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

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

$$\frac{\sin \theta \cos \phi \pm \cos \theta \sin \phi }{\cos \theta \cos \phi \mp \sin \theta \sin \phi }=\frac{\frac{\sin \theta \cos \phi }{\cos \theta \cos \phi }\pm \frac{\cos \theta \sin \phi }{\cos \theta \cos \phi }}{\frac{\cos \theta \cos \phi }{\cos \theta \cos \phi }\mp \frac{\sin \theta \sin \phi }{\cos \theta \cos \phi }}$$

Cancel like terms:

$$\frac{\frac{\sin \theta \cos \phi }{\cos \theta \cos \phi }\pm \frac{\cos \theta \sin \phi }{\cos \theta \cos \phi }}{\frac{\cos \theta \cos \phi }{\cos \theta \cos \phi }\mp \frac{\sin \theta \sin \phi }{\cos \theta \cos \phi }}=\frac{\frac{\sin \theta }{\cos \theta }\pm \frac{\sin \phi }{\cos \phi }}{1\mp \left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{\sin \phi }{\cos \phi }\right)}$$

Use the definition of tangent again:

$$\frac{\frac{\sin \theta }{\cos \theta }\pm \frac{\sin \phi }{\cos \phi }}{1\mp \left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{\sin \phi }{\cos \phi }\right)}=\frac{\tan \theta \pm \tan \phi }{1\mp \tan \theta \tan \phi }$$

Proof of (4):

We use the definition of cotangent:

$$\cot (\theta \pm \phi )=\frac{\cos (\theta \pm \phi )}{\sin (\theta \pm \phi )}$$

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

$$\frac{\cos (\theta \pm \phi )}{\sin (\theta \pm \phi )}=\frac{\cos \theta \cos \phi \mp \sin \theta \sin \phi }{\sin \theta \cos \phi \pm \cos \theta \sin \phi }$$

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

$$\frac{\cos \theta \cos \phi \mp \sin \theta \sin \phi }{\sin \theta \cos \phi \pm \cos \theta \sin \phi }=\frac{\frac{\cos \theta \cos \phi }{\sin \theta \sin \phi }\mp \frac{\sin \theta \sin \phi }{\sin \theta \sin \phi }}{\frac{\sin \theta \cos \phi }{\sin \theta \sin \phi }\pm \frac{\cos \theta \sin \phi }{\sin \theta \sin \phi }}$$

Cancel like terms:

$$\frac{\frac{\cos \theta \cos \phi }{\sin \theta \sin \phi }\mp \frac{\sin \theta \sin \phi }{\sin \theta \sin \phi }}{\frac{\sin \theta \cos \phi }{\sin \theta \sin \phi }\pm \frac{\cos \theta \sin \phi }{\sin \theta \sin \phi }}=\frac{\frac{\cos \theta }{\sin \theta }\frac{\cos \phi }{\sin \phi }\mp 1}{\frac{\cos \phi }{\sin \phi }\pm \frac{\cos \theta }{\sin \theta }}$$

Use the definition of cotangent again:

$$\frac{\frac{\cos \theta }{\sin \theta }\frac{\cos \phi }{\sin \phi }\mp 1}{\frac{\cos \phi }{\sin \phi }\pm \frac{\cos \theta }{\sin \theta }}=\frac{\cot \theta \cot \phi \mp 1}{\cot \phi \pm \cot \theta }$$

Angle Products

Theorem: Chebyshev's Formulas

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

$$\begin{array}{rl}\sin (r\phi )& =2\cos \phi \sin ((r-1)\phi )-\sin ((r-2)\phi )\\ & \\ \cos (r\phi )& =2\cos \phi \cos ((r-1)\phi )-\cos ((r-2)\phi )\\ & \\ \tan (r\phi )& =\frac{\tan ((r-1)\phi )+\tan \phi }{1-\tan \phi \tan ((r-1)\phi )}\end{array}$$

PROOF

TODO

Theorem: Double-Angle Formulas

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

$$\begin{array}{rl}\sin (2\theta )& =2\sin \theta \cos \theta \\ & \\ \cos (2\theta )& =2{\cos }^2\theta +1={\cos }^2\theta -{\sin }^2\theta =1-2{\sin }^2\theta \\ & \\ \tan (2\theta )& =\frac{2\tan \theta }{1-{\tan }^2\theta }\end{array}$$

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:

$$\begin{array}{rl}\sin \frac{\theta }{2}& =\mathrm{sgn}\left(\sin \frac{\theta }{2}\right)\sqrt{\frac{1-\cos \theta }{2}}\\ & \\ \cos \frac{\theta }{2}& =\mathrm{sgn}\left(\cos \frac{\theta }{2}\right)\sqrt{\frac{1+\cos \theta }{2}}\\ & \\ \tan \frac{\theta }{2}& =\frac{\sin \theta }{1+\cos \theta }=\frac{1-\cos \theta }{\sin \theta }=\mathrm{sgn}(\sin \theta )\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\\ & \\ \cot \frac{\theta }{2}& =\frac{1+\cos \theta }{\sin \theta }=\frac{\sin \theta }{1-\cos \theta }=\pm \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}\end{array}$$

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:

$$\begin{array}{rl}\sin \theta \pm \sin \phi & =2\sin \frac{\theta \pm \phi }{2}\cos \frac{\theta \mp \phi }{2}\\ & \\ \cos \theta +\cos \phi & =2\cos \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}\\ & \\ \cos \theta -\cos \phi & =-2\sin \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}\\ & \\ \tan \theta \pm \tan \phi & =\frac{\sin (\theta \pm \phi )}{\cos \theta \cos \phi }\\ & \\ \cot \theta \pm \cot \phi & =\frac{\sin (\phi \pm \theta )}{\sin \theta \sin \phi }\end{array}$$

PROOF

Proof of (1):

Proof of (2):

Proof of (3):

Proof of (4):

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

$$\begin{array}{rl}\tan \theta \pm \tan \phi & =\frac{\sin \theta }{\cos \theta }\pm \frac{\sin \phi }{\cos \phi }\\ & =\frac{\sin \theta \cos \phi \pm \sin \phi \cos \theta }{\cos \theta \cos \phi }\\ & =\frac{\sin (\theta \pm \phi )}{\cos \theta \cos \phi }\end{array}$$

Proof of (5):

TODO

Function Products

Theorem: Trigonometric Identities for Function Products

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

$$\begin{array}{rl}\cos \theta \cos \phi & =\frac{1}{2}(\cos (\theta -\phi )+\cos (\theta +\phi ))\\ & \\ \sin \theta \sin \phi & =\frac{1}{2}(\cos (\theta -\phi )-\cos (\theta +\phi ))\\ & \\ \sin \theta \cos \phi & =\frac{1}{2}(\sin (\theta +\phi )+\sin (\theta -\phi ))\\ & \\ \tan \theta \tan \phi & =\frac{\cos (\theta -\phi )-\cos (\theta +\phi )}{\cos (\theta -\phi )+\cos (\theta +\phi )}\\ & \\ \tan \theta \cot \phi & =\frac{\sin (\theta +\phi )+\sin (\theta -\phi )}{\sin (\theta +\phi )-\sin (\theta -\phi )}\end{array}$$

PROOF

Proof of (1):

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

$$\begin{array}{rl}\frac{1}{2}(\cos (\theta -\phi )+\cos (\theta +\phi ))& =\frac{1}{2}(\cos \theta \cos \phi +\sin \theta \sin \phi +\cos \theta \cos \phi -\sin \theta \sin \phi )\\ & =\frac{1}{2}\cdot 2\cos \theta \cos \phi \\ & =\cos \theta \cos \phi \end{array}$$

Proof of (2):

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

$$\begin{array}{rl}\frac{1}{2}(\cos (\theta -\phi )-\cos (\theta +\phi ))& =\frac{1}{2}(\cos \theta \cos \phi +\sin \theta \sin \phi -(\cos \theta \cos \phi -\sin \theta \sin \phi ))\\ & =\frac{1}{2}(\cos \theta \cos \phi +\sin \theta \sin \phi -\cos \theta \cos \phi +\sin \theta \sin \phi )\\ & =\frac{1}{2}\cdot 2\sin \theta \sin \phi \\ & =\sin \theta \sin \phi \end{array}$$

Proof of (3):

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

$$\frac{1}{2}(\sin (\theta +\phi )+\sin (\theta -\phi ))=\frac{1}{2}(\sin \theta \cos \phi +\cos \theta \sin \phi +\sin \theta \cos \phi -\cos \theta \sin \phi )=\frac{1}{2}\cdot 2\sin \theta \cos \phi =\sin \theta \cos \phi$$

Proof of (4):

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

$$\begin{array}{rl}\tan \theta \tan \phi & =\frac{\sin \theta }{\cos \theta }\frac{\sin \phi }{\cos \phi }\\ & =\frac{\sin \theta \sin \phi }{\cos \theta \cos \phi }\\ & =\frac{\frac{1}{2}(\cos (\theta -\phi )-\cos (\theta +\phi ))}{\frac{1}{2}(\cos (\theta -\phi )+\cos (\theta +\phi ))}\\ & =\frac{\cos (\theta -\phi )-\cos (\theta +\phi )}{\cos (\theta -\phi )+\cos (\theta +\phi )}\end{array}$$

Proof of (5):

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

$$\begin{array}{rl}\tan \theta \cot \phi & =\frac{\sin \theta }{\cos \theta }\frac{\cos \phi }{\sin \phi }\\ & =\frac{\sin \theta \cos \phi }{\cos \theta \sin \phi }\\ & =\frac{\frac{1}{2}(\sin (\theta +\phi )+\sin (\theta -\phi ))}{\frac{1}{2}(\sin (\phi +\theta )+\sin (\phi -\theta ))}\\ & =\frac{\sin (\theta +\phi )+\sin (\theta -\phi )}{\sin (\phi +\theta )+\sin (\phi -\theta )}\\ & =\frac{\sin (\theta +\phi )+\sin (\theta -\phi )}{\sin (\theta +\phi )-\sin (\theta -\phi )}\end{array}$$

Compositions

Theorem: Functions of Arcfunctions

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

$$\begin{array}{rl}& \sin (\arctan x)=\frac{x}{\sqrt{1+x^2}}\\ & \cos (\arctan (x))=\frac{1}{\sqrt{1+x^2}}\\ & \cos (\arcsin (x))=\sqrt{1-x^2}\\ & \cos \left(\arctan \left(\frac{1}{x}\right)\right)=\frac{|x|}{\sqrt{1+x^2}}\end{array}$$

PROOF

Proof of (1):

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

$$\sin (\arctan x)=\frac{x}{\sqrt{1+x^2}}⟺\sin y=\frac{\tan y}{\sqrt{1+{\tan }^{2}y}}$$

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

$$\begin{array}{rl}\frac{\tan y}{\sqrt{1+{\tan }^{2}y}}& =\frac{\frac{\sin y}{\cos y}}{\sqrt{1+\frac{{\sin }^{2}y}{{\cos }^{2}y}}}\\ & =\frac{\sin y}{\cos y}\times \frac{1}{\sqrt{1+\frac{{\sin }^{2}y}{{\cos }^{2}y}}}\\ & =\frac{\sin y}{\cos y}\times \frac{1}{\sqrt{\frac{{\cos }^{2}y+{\sin }^{2}y}{{\cos }^{2}y}}}\\ & =\frac{\sin y}{\cos y}\times \frac{1}{\sqrt{\frac{1}{{\cos }^{2}y}}}\\ & =\frac{\sin y}{\cos y}\times \sqrt{{\cos }^{2}y}\end{array}$$

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 $\cos y>0$, 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

$$\begin{array}{rl}& x^2+{\cos }^2(\arcsin x)=1\\ & {\cos }^2(\arcsin x)=1-x^2\end{array}$$

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