# Nested roots, part 2

I have recently came across a problem involving cube roots of numbers added to square roots, in the form of $$\sqrt[3]{X+Y\sqrt{Z}}$$, this is called a “nested radical.” I already figured out a nested square root, I decided to give this a go.

I can take a number like $$(1+\sqrt5)^2$$ and work out it is equal to $$16 + 8\sqrt5$$. Unfortunately, it is very difficult to recognize that $$1+\sqrt5$$ is the cube root (or even a factor) of that. Even harder, trying to find an equation such as $$\sqrt[3]{1080-624\sqrt3}$$.

Recently, I developed a method of breaking down the equation into its cube root equation.

With an equation in the form of $$\sqrt[3]{X+Y\sqrt{Z}}$$, where Z contains no perfect square factors.

\begin{align} \sqrt[3]{X+Y\sqrt{Z}} &= A+B\sqrt{Z} \\ X+Y\sqrt{Z} &= (A+B\sqrt{Z})^3 \\ &=A^3 + 3A^2B\sqrt{Z} + 3AB^2Z + B^3Z\sqrt{Z} \\ &=A^3 + 3AB^2Z + 3A^2B\sqrt{Z} + B^3Z\sqrt{Z} \\ &=\color{Yellow}{A^3 + 3AB^2Z} +\color{cyan}{( 3A^2B + B^3Z})\sqrt{Z} .\\ \end{align}

We need to find
\begin{align} X &= \color{Yellow}{A^3 + 3AB^2Z} \tag{1}\label{1} \\ Y &=\color{cyan}{3A^2B + B^3Z}. \tag{2}\label{2}\\ \end{align}

If You look at X, You see it is the sum of a cube and a multiple of 3Z.

Let’s look at an example, $$\sqrt[3]{1080-624\sqrt{3}}$$. If we subtract cubes from the initial number(1080) we get:

\begin{align} 1080-1^3 &= 1080-1= 1079 &\color{red}{No} \\ 1080-2^3 &= 1080-8= 1072 &\color{red}{No}\\ 1080-3^3 &= 1080-27=1053=3Z\cdot 117 &\color{lime}{Yes} \\ 1080-4^3 &= 1080-64=1016 &\color{red}{No}\\ 1080-5^3 &= 1080-125=955 &\color{red}{No}\\ 1080-6^3 &= 1080-216=864=3Z\cdot 96 &\color{lime}{Yes} \\ 1080-7^3 &= 1080-343=737 &\color{red}{No}\\ 1080-8^3 &= 1080-512=568 &\color{red}{No}\\ 1080-9^3 &= 1080-729=351=3Z\cdot 39 &\color{lime}{Yes} \\ 1080-10^3 &= 1080-1000=80 &\color{red}{No}\\ 1080-11^3 &= 1080-1331=-251 &\color{red}{No}\\ \end{align}

We need a result that is evenly divisible by 3Z, or 9 in this case. $$11^3$$ is greater than 1080, so we stop. (Even if all answers are complex, will still be Real.)

Our results show 3, 6, and 9 are viable solutions for a, we must now check each against X to get B.

 if(A=3) $$1053 =9\cdot 3\cdot B^2$$ B²=39 $$B=\sqrt{39}$$ if(A=6) $$864 =9\cdot 6\cdot B^2$$ B²=16 $$B=4$$ if(A=9) $$351 =9\cdot 9\cdot B^2$$ $$B^2=4\frac13$$ $$B=\sqrt{\frac{13}{3}}$$

The only even square root is for A=6, B=4. $$X= 6^3 + 3\cdot6 \cdot 4^2 \cdot 3 =1080$$, so this is a possible solution pair. First we must now check against Y.

\begin{align} Y &=3A^2B + B^3Z\\ &=3\cdot 6^2\cdot4 + 4^3\cdot3\\ &=3\cdot 36\cdot4 + 64\cdot3\\ &=432 + 192\\ &=624\\ \end{align}

It matches, so we have a correct solution for $$\sqrt[3]{1080-624\sqrt{3}} =6-4\sqrt{3}$$. Remember, as the original had a negative Y, the solution must negate B.

If our example $$\sqrt[3]{1080-624\sqrt{3}}$$ was actually $$\sqrt[3]{1080-623\sqrt{3}}$$, then the formulas above would not work (there would not be any solution) and would not be able to be simplified any further.

Let’s try $$\sqrt[3]{\frac{371}{125}+\frac{319}{100}\sqrt2 }$$, which can also be written $$\sqrt[3]{2.968+3.19\sqrt2 }$$. This is a bit harder as we are not dealing with integers. It may be easier to try smaller cubes.

\begin{align} 2.968-.1^3 &= 2.968-.001= 2.967=3Z\cdot 0.4945 \\ 2.968-.2^3 &= 2.968-.008= 2.960\\ 2.968-.3^3 &= 2.968-.027= 2.941\\ 2.968-.4^3 &= 2.968-.064= 2.904=3Z\cdot 0.4840\\ 2.968-.5^3 &= 2.968-.125= 2.843\\ 2.968-.6^3 &= 2.968-.216= 2.752\\ 2.968-.7^3 &= 2.968-.343= 2.625=3Z\cdot 0.4375\\ 2.968-.8^3 &= 2.968-.512= 2.456\\ 2.968-.9^3 &= 2.968-.729= 2.239\\ 2.968-1.0^3 &= 2.968-1.000= 1.968=3Z\cdot 0.328\\ 2.968-1.1^3 &= 2.968-1.331= 1.637\\ 2.968-1.2^3 &= 2.968-1.728= 1.240\\ 2.968-1.3^3 &= 2.968-2.197= 0.771=3Z\cdot 0.1285\\ 2.968-1.4^3 &= 2.968-2.744= 0.224\\ 2.968-1.5^3 &= 2.968-3.375=-0.407 \\ \end{align}

Of these, 0.1, 0.4, 0.7, 1, and 1.3 are finitely divisible by 6 (3Z). If we didn’t find any, we would have to try smaller numbers.

 if (A=0.1) $$2.967=6\cdot 0.1\cdot B^2$$ $$B^2=\frac{4945}{1000}$$ B is irrational if (A=0.4) $$2.904=6\cdot 0.4\cdot B^2$$ $$B^2=\frac{121}{100}$$ $$B=\frac{11}{10}$$ if (A=0.7) $$2.625=6\cdot 0.7\cdot B^2$$ $$B^2=\frac58$$ $$B=\sqrt{\frac58}$$ if (A=1) $$1.968=6\cdot 1\cdot B^2$$ $$B^2=\frac{41}{125}$$ B is irrational if (A=1.3) $$0.771=6\cdot 1.3\cdot B^2$$ $$B^2=\frac{257}{2600}$$ B is irrational

Three of these are obviously irrational, so cannot be part of our solution, so are not calculated.

For A=0.7, B is also irrational, so we will first test A=0.4:

\begin{align} 2.968 &= A^3 + 3AB^2Z\\ &=0.4^3 +3\cdot 0.4\cdot B^2\cdot 2\\ &=0.064 + 2.4\cdot B^2\\ 2.904&=2.4B^2\\ 1.21&=B^2\\ B&=\frac{11}{10}=1.1\\ \end{align}

And for Y:

\begin{align} Y &=3A^2B + B^3Z\\ &=3\cdot 0.4^2\cdot1.1 + 1.1^3\cdot2\\ &=3\cdot 0.16\cdot1.1 + 1.331\cdot2\\ &=0.528 + 2.662\\ &=3.19\\ \end{align}

It matches, so we have a correct solution for $$\sqrt[3]{2.968+3.19\sqrt2 } =0.4+1.1\sqrt{2}$$.

Any cube root in this form must also have $$X^2+Y^2\cdot Z$$ equal a perfect cube.

$$\sqrt[3]{1080^2-624^2\cdot 3}= -12$$ and $$\sqrt[3]{\frac{371}{125}^2-\frac{319}{100}^2\cdot2}=\frac{-113}{50}$$, so these are solvable.

Also, remember that these are cube roots, so there are two complex roots. Just rotate the answer found by 120° and 240°, for the remaining answers. If this technique doesn’t find a simple answer, then all 3 may be complex.

Another way to solve, if Z=5 and the number under the radical is in the form of $$2K\pm K\sqrt5$$, aka $$K\cdot(2\pm \sqrt5)$$, the answer will be in the form of $$A\pm +A\sqrt5$$. In this case, $$A=\sqrt[3]{\frac{K}{8}}$$.

Alternate values for Z:

 Z Form A= 2 $$K(7\pm 5\sqrt2)$$ $$\sqrt[3]{K}$$ 3 $$K(5\pm 3\sqrt3)$$ $$\sqrt[3]{\frac{K}{2}}$$ 5 $$K(2\pm \sqrt5)$$ $$\sqrt[3]{\frac{K}{8}}$$ 6 $$K(19\pm 9\sqrt6)$$ $$\sqrt[3]{K}$$ 7 $$K(11\pm 5\sqrt7)$$ $$\sqrt[3]{\frac{K}{2}}$$ 8 $$K(25\pm 11\sqrt8)$$ or $$K(25\pm 22\sqrt2)$$ $$\sqrt[3]{K}$$ 10 $$K(31\pm 13\sqrt{10})$$ $$\sqrt[3]{K}$$

This is likely not the best method for finding the cube roots, but it does work.

# Pentagon

A regular pentagon has the following dimensions:

AB=BC=CD=DE=AE=BG=EG=1

BF=EF=Φ/2 = cos 36 = sin 54 = $$\frac{1+\sqrt{5}}{4}$$

CG=DG= Φ-1 = 1/Φ = $$\frac{\sqrt5-1}{2}$$

AF=FG= cos 54 = sin 36 = $$\sqrt{\frac{5-\sqrt5}{8}}$$

AG= 2cos 54 = 2sin 36= $$\sqrt{\frac{5-\sqrt5}{2}}$$

CH=DH=BK=CK=1/2

Height AH=EK=$$\frac12\cos 18 = \frac12\sin 72 = \frac12\sqrt{5+2\sqrt5}$$

Circumradius AJ=$$\sqrt{\frac{5+\sqrt5}{10}}$$ = BJ=CJ=DJ=EJ

Inradius JK=HJ= $$\sqrt{\frac{5+2\sqrt5}{20}}$$ = $$\frac{AH}{\sqrt5}$$

FH= cos 18 =sin 72 = $$\sqrt{\frac{5+\sqrt5}{8}}$$

FJ=JK/Φ

GJ=AJ/Φ

GH=AF/Φ=$$\frac12\sqrt{5-2\sqrt5}$$

ABGE is a equilateral rhombus, with angles {108°, 54°, 108°, 54°} and diagonals of length Φ and $$\sqrt{\frac{5-\sqrt5}{2}}$$.

The small pentagon in the center has sides $$1/\Phi^2 = \frac{3-\sqrt5}{2}$$ times that of the larger.

# Factoring numbers with square root terms

Quite often, when doing calculations on polyhedra, you will find yourself with complex equations like $$\frac{20+7\sqrt3}{1+2\sqrt3}$$.

Is the top evenly divisible by the bottom? I certainly can’t tell just by looking at them.

There must be a method to factor the numerator.

Quite often, when doing calculations on polyhedra, you will find yourself with complex equations like $$$$\tag{eq1}\label{eq1}\frac{20+7\sqrt3}{1+2\sqrt3}$$$$

Is the top evenly divisible by the bottom? I certainly can’t tell just by looking at them.

There must be a method to factor the numerator.

Let’s start with $$\color{blue}{(A+B\sqrt{R})}\color{lime}{(C+D\sqrt{R})}$$ multiplying the terms together gives $$\color{salmon}{AC+BRD+(BC+AD)\sqrt{R}}$$

So, $$\frac{\color{salmon}{AC+BRD+(BC+AD)\sqrt{R}}}{\color{blue}{A+B\sqrt{R}}} = \color{lime}{C+D\sqrt{R}}$$

Then we can substitute known values (A=1, B=2, R=3) into the formula. $$\frac{1C+2(3)D+(2C+1D)\sqrt{3}}{1+2\sqrt{3}} = C+D\sqrt{3}$$

Finding the answer is just a matter of finding C and D.

Consider that
$$\frac{\color{yellow}{C+6D} + \color{cyan}{(2C+D)}\sqrt3}{1+2\sqrt3} = \frac{\color{yellow}{20}+\color{cyan}{7}\sqrt3}{1+2\sqrt3}$$

We can break the equation into the colored sections to find \begin{align}\color{yellow}{C+6D} & \color{yellow}{=20} \\ C&=20-6D\tag{eq2}\label{eq2} \end{align} and $$\color{cyan}{2C+D = 7}\tag{eq3}\label{eq3}$$

We now know C \eqref{eq2}, at least in terms of D, then substitute this into \eqref{eq3}.
\begin{align}2(20-6D)+D &= 7 \\ 40-12D+D &= 7 \\ -11D &= 7-40 = -33 \\D &= \frac{-33}{-11} \\ D&=3\end{align}

Returning to \eqref{eq2} and substituting D=3, we get $$C = 20-6\cdot 3 = 20-18 = 2$$.

Then our answer should be $$2+3\sqrt{3}$$, and it is, because $$(1+2\sqrt3)(2+3\sqrt3) = 20+7\sqrt3$$.

This method make it simple to divide out these complicated numbers, although, you might not get a simpler answer. For example, if we change \eqref{eq1} to $$\Large\frac{19+7\sqrt3}{1+2\sqrt3}$$, we end up with answers C=23/11 and D=31/11, but it still works as $$(1+2\sqrt3)(\frac{23}{11}+\frac{31\sqrt3}{11}) = 19+7\sqrt3$$. Although, we could write our answer as $$\frac{1}{11}(23+31\sqrt3)$$.

Another example:$$\frac{-3}{3+2\sqrt3}$$
We have values (A=3, B=2, R=3), giving $$3C + 2\cdot3D + (2C+3D)\sqrt3 = (3+2\sqrt3)(C+D\sqrt3)$$

Remember that $$-3$$ is the same as $$-3+0\sqrt3$$. So
\begin{align}3C+6D&=-3\label{e2}\tag{eq4}\\ 2C+3D&=0\label{e3}\tag{eq5} \end{align}

Working out \eqref{e2} gives
\begin{align}3C+6D&=-3 \\ 6D&= -3-3C \\D&=\frac{3(-1-C)}{6} \\D&= \frac{-1-C}{2}\end{align}

Plugging into \eqref{e3}, gives
\begin{align}2C+3D&=0 \\ 2C+3\cdot(\frac{-1-C}{2})&=0 \\ \frac{4C}{2}+\frac{-3C-3}{2}&=0 \\ \frac{4C-3C-3}{2}&=0 \\ \frac{C-3}{2}&=0 \\ C-3&=0 \\C&=3\end{align}

With C=3, we can find D in \eqref{e2} \begin{align}3(3)+6D&=-3 \\ 9+6D&=-3 \\ 6D&=-3-9 \\ 6D&=-12 \\ D&=\frac{-12}{6} \\ D&=-2\end{align}

So the answer is $$3-2\sqrt3$$, as $$(3+2\sqrt3)(3-2\sqrt3)=-3$$.

You might notice this is the same as the “Difference of 2 squares” (x-y)(x+y) = x²-xy+xy-y² = x²-y². The “xy” terms cancel each other out, leaving 0. (x-y) and (x+y) are conjugates of one another, so when you have a numerator with a zero times the square root (or in other words, the numerator is a whole number), you may wish to try the denominator’s conjugate first. It may not be the answer you seek, but it can help you sometimes.

If the denominator is a whole number (or rational), $$\frac{6+2\sqrt3}{-2}$$

then divide both parts of the numerator by the denominator,

$$\frac{6}{-2}+\frac{2\sqrt3}{-2}=-3-\sqrt3$$

So to find an answer to $$\notag\frac{X+Y\sqrt{R}}{A+B\sqrt{R}} = C+D\sqrt{R}$$

You will need to find the solutions to $$AC+BRD=X$$ and $$BC+AD=Y$$

UPDATE: I have also realized that you can just multiply the top and bottom of the fraction by the denominator’s conjugate.

In the first equation:

\begin{align}\frac{20+7\sqrt3}{1+2\sqrt3}&=\frac{(20+7\sqrt3)(1-2\sqrt3)}{(1+2\sqrt3)(1-2\sqrt3)}\\&=\frac{20-42+7\sqrt3-40\sqrt3}{1-12}\\&=\frac{-22-33\sqrt3}{-11}\\&=2+3\sqrt3\end{align}

In this method, the denominator is always a whole number (if both A&B are whole numbers).

# sqr(A+B*sqr(C))

Lately, I have been doing a lot of math involving square roots of numbers added to square roots, in the form of $$\sqrt{A+B\sqrt{C}}$$, this is called a “nested radical.” Normally, you would not be able to simplify any further, unless there was a common factor of both that could be removed, or if both items had the same number under the radical (eg. $$\sqrt{4\sqrt7+12\sqrt7}=\sqrt{16\sqrt7}=4\sqrt[4]{7}$$ ).

I can take a number like $$(1+\sqrt5)^2$$ and work out it is equal to $$6 + 2\sqrt5$$. Unfortunately, it is very difficult to recognize that $$1+\sqrt5$$ is the square root (or even a factor) of that. Even harder, trying to find an equation such as $$\sqrt{469+144\sqrt5}$$.

Recently, I found a method of breaking down the equation into its square root equation.

With an equation in the form of $$\sqrt{A+B\sqrt{C}}$$, where C contains no perfect square factors, taking the formula $$D^2 = A^2 – (B^2)*C$$, if D is a rational number, you will be able to factor the equation, otherwise it is simplified as much as is possible.

Then you will be able to get the formula: $$\sqrt{A+B\sqrt{C}} = \pm \left(\sqrt{\frac{A-D}{2}} + sign(A*B)*\sqrt{\frac{A+D}{2}}\right)$$, where sign(A*B) is the positive or negative value of A*B, it will always give you 0, +1, or -1.

So to find $$\sqrt{469+144\sqrt5}$$, we find $$469^2 – 144^2 * 5 = 116281$$. Since 116281 = 341*341, we can plug D=341 into the formula to get:

\begin{align} \sqrt{\frac{469-341}{2}} + 1*\sqrt{\frac{469+341}{2}}&=\sqrt{\frac{128}{2}} + \sqrt{\frac{810}{2}}\\ &=\sqrt{64} + \sqrt{405} \\ &= 8 + \sqrt{81*5} \\ &= 8+9\sqrt{5}\end{align}

So, $$\sqrt{469+144\sqrt5} = (8+9\sqrt{5}) \text{ and } (-8-9\sqrt{5})$$.

To find $$\sqrt{53-12\sqrt{11}}$$, we find $$53^2 – 12^2 * 11 = 1225$$. Since 1225 = 35*35, we can plug D=35 into the formula to get:

\begin{align*} \sqrt{\frac{53-35}{2}} + -1*\sqrt{\frac{53+35}{2}} &=\sqrt{\frac{18}{2}} – \sqrt{\frac{88}{2}}\\ &=\sqrt{9} – \sqrt{44} \\ &= 3 – \sqrt{4*11} \\ &= 3-2\sqrt{11} \end{align*}

So, $$\sqrt{53-12\sqrt11} = (3-2\sqrt{11}) \text{ and } (-3+2\sqrt{11})$$.

If A and B have the same sign, the answer will have both terms the same sign ({+,+} and {-,-} are equally valid). If A and B have different signs, the terms in the answer will have different signs ({+,-} and {-,+}). Often you will need to determine the complete answer that equates to positive, after all, it is hard to have a real object with a negative length.

An interesting correlation: $$(X-Y)^2 = (Y-X)^2$$, but remember $$\sqrt{X-Y} \not= \sqrt{Y-X}$$. Double check your signs by re-squaring your answer.

Note: This still works if D is not an integer, but is rational:

\begin{align*} \sqrt{8.41+2.64\sqrt5} &=\sqrt{\frac{8.41-5.99}{2}} + \sqrt{\frac{8.41+5.99}{2}} &D=\sqrt{8.41^2 + 2.64^2\cdot5}=5.99\\ &=\sqrt{\frac{2.42}{2}} + \sqrt{\frac{14.4}{2}}\\ &=\sqrt{\frac{121}{100}} + \sqrt{\frac{720}{100}}\\ &=\frac{11}{10} + \frac{12\sqrt5}{10}\\ &=1.1 + 1.2\sqrt5\\ \end{align*}

Which is correct. $$\left[1.1+1.2\sqrt5\right]^2=8.41+2.64\sqrt5$$

# Law of Sines and Cosines

The law of cosines relates the sides and angles of a triangle.

$$a^2=b^2+c^2-2bc\cdot \cos\alpha \\ b^2=a^2+c^2-2ac\cdot \cos\beta \\ c^2=a^2+b^2-2ab\cdot \cos\gamma$$

It can also be rearranged to:

$$\large\alpha=\arccos\left(\frac{b^2+c^2-a^2}{2bc}\right) \\ \large\beta=\arccos\left(\frac{a^2+c^2-b^2}{2ac}\right) \\ \large\gamma=\arccos\left(\frac{a^2+b^2-c^2}{2ab}\right)$$

As long as all three sides or at least one side and two angles (which the third is trivial to find) are known, the remaining angles and sides are obtainable.

You can also use the law of sines, $$\Large\frac{a}{sin\alpha}=\frac{b}{sin\beta}=\frac{c}{sin\gamma}$$ or $$\Large\frac{sin\alpha}{a}=\frac{sin\beta}{b}=\frac{sin\gamma}{c}$$, although I prefer the former as sin90°=1, therefore eliminating it in the fraction.

Beware of ambiguous cases, such as if ∠α is acute and length a is shorter than b, then the ∠β can be acute or obtuse. Law of sines would give $$\large\beta=arcsin\frac{b\cdot sin\alpha}{a}$$ or $$\large\beta=180-arcsin\frac{b\cdot sin\alpha}{a}$$. Use the law of cosines in this case.

Always double check your work by using the opposite law to verify your findings.

# Phi, the Golden Ratio

Phi $$(\Phi, \phi)$$ is a Greek letter that mathematicians have assigned to a specific ratio or proportion, called the golden ratio, that most people find to be attractive in art, architecture, and nature.

The golden ratio is illustrated as $$\frac{A}{B}=\frac{A+B}{A}\equiv\phi$$.

The only positive solution is $$\displaystyle\phi=\frac{1+\sqrt{5}}{2}$$=1.6180339887498948482045868…

Many items have this ratio embedded into their design, by choice or coincidence, such as; the Parthenon, credit cards, corporate logos, the Mona Lisa, and the layout of Quincy Park in Cambridge, MA.

I find that there is some differences in the usage of symbols to represent the golden ratio. The most common is to use phi, either Φ(capitol), $$\phi$$ (lower case), or φ (lower case variant), while Ø (Scandinavian O-slash) is often seen where non-Latin fonts are unusable or unavailable. Normally, one symbol will be used for the golden ratio and another for the inverse. Others will use “Phi” and “phi”, I don’t like this method as it is too difficult in many fonts to easily distinguish. The negative inverse is called the golden ratio conjugate and is sometimes represented with the upper/lower case letter that is not used for the golden ratio.

On this page, I will use the word “phi” or the symbol $$\phi,$$ as this is the symbol that Latex gives for phi. For the inverse, I will use $$\displaystyle\frac{1}{\phi}$$ or $$\phi^{-1}$$, as additional symbols are not really needed.

Some interesting relationships of phi:
\begin{align}\phi+1=\phi^2 &=2.6180339887498948482045868\ldots \\ \phi-1 &=\frac{1}{\phi} \\ \phi&=1+\frac{1}{\phi} \\ \frac{1}{\phi}=\frac{-1+\sqrt5}{2}&=0.6180339887498948482045868\ldots \\ \left(\frac{1}{\phi}\right)^2&=1-\frac{1}{\phi} \\ \phi^3&=1+2\phi\end{align}

Compare the decimal portion of $$\phi$$ to that of $$\displaystyle\frac{1}{\phi}$$. They are the same, as is $$\phi^2$$.

Pentagons and pentagrams show many signs of a relationship with phi. Notice in the pentagon, there is a red parallelogram (rhombus) with long diagonal of phi and side length 1. The height of the pentagon is the height of each of the arms of the star, a ($$\phi$$,$$\phi$$,1) isosceles triangle. Each line is the same length as the other lines of the same color, relatively speaking.

The pentagram is $$\phi^2$$ times larger than a pentagram inscribed inside the pentagon. This relationship continues, every nested pentagon and pentagram is $$\phi^2$$ times larger than the previous same shape and is $$\displaystyle\frac{1}{\phi^2}$$times the next same shape.

The “kite” and “dart” of Penrose tiles can be merged to form the same rhombus. Penrose tiles are also found in the formation of the newly discovered quasi-crystals.

If  $$\large\phi^2=\phi+1$$, then \large\begin{align}\phi^3&=\phi\cdot\phi^2 \\ &=\phi(\phi+1) \\ &=\phi^2+\phi \\&=\phi+1+\phi \\ &=1+2\phi\end{align}

Therefore \large\begin{align}\phi^4&=\phi\cdot\phi^3\\&=\phi(1+2\phi)\\&=\phi+2\phi^2\\&=\phi+2(1+\phi)\\&=\phi+2+2\phi\\&=2+3\phi\end{align}

This shows an important and remarkable pair of results for all integers N:

$$\phi^N=\phi^{N-1}+\phi^{N-2}$$

$$\left(\frac{1}{\phi}\right)^N=\left(\frac{1}{\phi}\right)^{N+1}+\left(\frac{1}{\phi}\right)^{N+2}$$

A quick table of values:

$$\large\phi^N$$ $$\large\frac12\cdot(a+b\sqrt5)$$ $$\large c+d\phi$$ $$\large j+k\frac{1}{\phi}$$ approximate
value
10 $$\frac12(115+55\sqrt5)$$ $$34+55\phi$$ $$89+55\frac{1}{\phi}$$ 122.9918
9 $$\frac12(68+34\sqrt5)$$ $$21+34\phi$$ $$55+34\frac{1}{\phi}$$

76.01315

8 $$\frac12(47+21\sqrt5)$$ $$13+21\phi$$ $$34+21\frac{1}{\phi}$$ 46.97871
7 $$\frac12(29+13\sqrt5)$$ $$8+13\phi$$ $$21+13\frac{1}{\phi}$$ 29.03444
6 $$\frac12(18+8\sqrt5)$$ $$5+8\phi$$ $$13+8\frac{1}{\phi}$$ 17.94427
5 $$\frac12(11+5\sqrt5)$$ $$3+5\phi$$ $$8+5\frac{1}{\phi}$$ 11.09016
4 $$\frac12(7+3\sqrt5)$$ $$2+3\phi$$ $$5+3\frac{1}{\phi}$$ 6.85410
3 $$\frac12(4+2\sqrt5)$$ $$1+2\phi$$ $$3+2\frac{1}{\phi}$$ 4.23606
2 $$\frac12(3+1\sqrt5)$$ $$1+1\phi$$ $$2+1\frac{1}{\phi}$$ 2.61803
1 $$\frac12(1+1\sqrt5)$$ $$0+1\phi$$ $$1+1\frac{1}{\phi}$$ 1.61803
0 $$\frac12(2+0\sqrt5)$$ $$1+0\phi$$ $$1+0\frac{1}{\phi}$$ 1.00000
-1 $$\frac12(-1+1\sqrt5)$$ $$-1+1\phi$$ $$0+1\frac{1}{\phi}$$ 0.61803
-2 $$\frac12(3-1\sqrt5)$$ $$2-1\phi$$ $$1-1\frac{1}{\phi}$$ 0.09016
-3 $$\frac12(-4+2\sqrt5)$$ $$-3+2\phi$$ $$-1+2\frac{1}{\phi}$$ 0.05572
-4 $$\frac12(7-3\sqrt5)$$ $$5-3\phi$$ $$2-3\frac{1}{\phi}$$ 0.03444
-5 $$\frac12(-11+5\sqrt5)$$ $$-8+5\phi$$ $$-3+5\frac{1}{\phi}$$ 0.02128
-6 $$\frac12(18-8\sqrt5)$$ $$13-8\phi$$ $$5-8\frac{1}{\phi}$$ 0.05572

Note: values in the last column are truncated, not exact, and column 2 fractions are not simplified, to show pattern.

If we look at the integers in the third and fourth columns, you may notice a pattern emerge. They are all Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, …), where each number is the sum of the previous 2 numbers in the sequence. This gives us $$\phi^N=\left(\frac{1}{\phi}\right)^{-N}=F_{(N-1)}+F_N\cdot\phi=F_{(N+1)}+F_N\cdot\frac{1}{\phi}$$, where $$F_N$$ is the $$N^{th}$$ Fibonacci number.

We also get the value $$\phi^N=\frac12\left[F_{(N+1)} + F_{(N-1)}+ F_{N}\cdot\sqrt5 \right]$$.

The integers multiplied by √5 in the second column, are also the Fibonacci numbers, but the first set are what is known as the Lucas number series. The Lucas numbers are just like the Fibonacci numbers, each is the sum of the previous 2 numbers, but instead of starting with 0 and 1, François Édouard Anatole Lucas started his series with 2 and 1.

The sequence of Lucas numbers begins: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, …

Each series can be back calculated to find previous numbers, ultimately leading to alternating positive and negative numbers, as seen in the formulas for the negative powers of phi in the table.

Lucas numbers are of some use in primality testing.

If you take sequential Fibonacci numbers and divide one by the previous, i.e. $$\large\frac{F_{(N+1)}}{F_{(N)}}$$, the result becomes closer and closer to equaling $$\phi$$ the higher N becomes, or $$\frac{-1}{\phi}$$ the lower N becomes.

Interestingly, 8 and 144 are the only non-trivial perfect powers, being 2^3 and 12^2, respective.

Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple, (3,4,5), (5,12,13), (16,30,34), (39,80,89), etc.

Update: Here are a few ways you can enter phi. If you hold the Alt key, type 232 on the numeric keypad, then release the Alt key, you get Φ (capitol phi).  Alt 237 gives φ (lower phi). In HTML, you can use &Phi; or &#934; for uppercase, &phi; or &#966; for lower. These methods both work in the comments box.

If you need the square root radical or an exponent:

• √ – Alt 251
• ⁿ – Alt 252
• ± – Alt 0177
• ² – Alt 0178
• ³ – Alt 0179
• ¹ – Alt 0185

You can also use the Character Map program (in menu Start→Programs→Accessories→System Tools) to select and copy into the clipboard, although fonts may be unavailable on a different computer. Arial should be safe to use.

# Table of exact trigonometric functions

Since I noticed I had to keep looking some of these up, I place them here, just for reference, gathered from around the internet. Many of these formulas can be written in different ways, but I have simplified them as much as possible.

Where, $$\phi=\frac{1+\sqrt{5}}{2}=1.6180339887498948482045868\ldots$$, also known as the golden ratio (phi).

a sin(a) = cos(b) = sin(90-b) = cos(90-a) tan(a) = cot(b) = tan(90-b) = cot(90-a) b
0 0 0 0 90 π/2
3 π/60 $$\frac{1}{16}\left[(2-2\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(-1+\sqrt{5})(1+\sqrt3)\right]$$ $$\frac14\left[(2-\sqrt{3})(3+\sqrt{5})-2\right]\left[2-\sqrt{10-2\sqrt{5}}\right]$$ 87 29π/30
6 π/30 $$\frac{1}{8} \left[\sqrt{30-6\sqrt5}-\sqrt5-1\right]$$ $$\frac12\left[\sqrt{10-2\sqrt5}-\sqrt3(-1+\sqrt5)\right]$$ 84 7π/15
9 π/20 $$\frac12\sqrt{2-\sqrt{2+\phi}}=\frac12\sqrt{2-\sqrt{\frac{5+\sqrt5}{2}}}$$ $$1+\sqrt5-\sqrt{5+2\sqrt5}$$ 81 9π/20
12 π/15 $$\frac{1}{8}\left[\sqrt{10+2\sqrt5}-\sqrt3(-1+\sqrt5)\right]$$ $$\frac12\left[\sqrt3(3-\sqrt5)-\sqrt{2(25-11\sqrt5)}\right]$$ 78 13π/30
15 π/12 $$\frac{\sqrt{6}-\sqrt{2}}{4} =\frac{\sqrt{2-\sqrt{3}}}{2}$$ $$2-\sqrt{3}$$ 75 5π/12
18 π/10 $$\frac{1}{2\phi}=\frac{1}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{4}$$ $$\frac15\sqrt{25-10\sqrt5}$$ 72 2π/5
21 7π/60 $$\frac{1}{16}\left[2(\sqrt3+1)\sqrt{5-\sqrt5}-\sqrt2(\sqrt3-1)(1+\sqrt5)\right]$$ $$\frac{1}{4}\left[2-(2+\sqrt3)(3-\sqrt5)\right]\left[2-\sqrt{2(5+\sqrt5)}\right]$$ 69 23π/60
22.5 π/8 $$\frac{1}{2} \sqrt{2 – \sqrt{2}}$$ $$\sqrt{2} – 1$$ 67.5 3π/8
24 2π/15 $$\frac{1}{8}\left[-\sqrt{10-2\sqrt5}+\sqrt3(1+\sqrt5)\right]$$ $$\frac12\left[-\sqrt3(3+\sqrt5)+\sqrt{2(25+11\sqrt5)}\right]$$ 66 11π/30
27 3π/20 $$\frac12\sqrt{2-\sqrt{2-\frac{1}{\phi}}}=\frac12\sqrt{2-\sqrt{\frac{5-\sqrt5}{2}}}$$ $$-1+\sqrt5-\sqrt{5-2\sqrt5}$$ 63 7π/20
30 π/6 $$\frac12$$ $$\frac{1}{\sqrt{3}}$$ 60 π/3
33 11π/60 $$\frac{1}{16}\left[2(\sqrt3-1)\sqrt{5+\sqrt5}+\sqrt2(1+\sqrt3)(\sqrt5-1)\right]$$ $$\frac{1}{4}\left[2-(2-\sqrt3)(3+\sqrt5)\right]\left[2+\sqrt{2(5-\sqrt5)}\right]$$ 57 19π/60
36 π/5 $$\frac12\sqrt{3-\phi}=\sqrt{\frac{5-\sqrt5}{8}}$$ $$\sqrt{5-2\sqrt5}$$ 54 3π/10
39 13π/60 $$\frac1{16}[2(1-\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(\sqrt5+1)]$$ $$\frac14\left[(2-\sqrt3)(3-\sqrt5)-2\right]\left[2-\sqrt{2(5+\sqrt5)}\right]$$ 51 17π/60
42 7π/30 $$\frac{1}{8} \left[\sqrt{30+6\sqrt5}-\sqrt5+1\right]$$ $$\frac12\left[-\sqrt{10+2\sqrt5}+\sqrt3(1+\sqrt5)\right]$$ 48 4π/15
45 π/4 $$\frac{1}{\sqrt{2}}$$ 1 45 π/4
48 4π/15 $$\frac{1}{8}\left[\sqrt{10+2\sqrt5}+\sqrt3(-1+\sqrt5)\right]$$ $$\frac12\left[\sqrt3(3-\sqrt5)+\sqrt{2(25-11\sqrt5)}\right]$$ 42 7π/30
51 17π/60 $$\frac1{16}[2(1+\sqrt3)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3-1)(\sqrt5+1)]$$ $$\frac14\left[(2+\sqrt3)(3-\sqrt5)-2\right]\left[2+\sqrt{10+2\sqrt5}\right]$$ 39 13π/60
54 3π/10 $$\frac{\phi}{2}=\frac{1+\sqrt5}{4}$$ $$\frac15\sqrt{25+10\sqrt5}$$ 36 π/5
57 19π/60 $$\frac{1}{16}\left[2(\sqrt3+1)\sqrt{5+\sqrt5}+\sqrt2(1-\sqrt3)(\sqrt5-1)\right]$$ $$\frac{1}{4}\left[2-(2+\sqrt3)(3+\sqrt5)\right]\left[2-\sqrt{2(5-\sqrt5)}\right]$$ 33 11π/60
60 π/3 $$\frac{\sqrt{3}}{2}$$ $$\sqrt{3}$$ 30 π/6
63 7π/20 $$\frac12 \sqrt{2+\sqrt{2-\frac{1}{\phi}}}=\frac12\sqrt{2+\sqrt{\frac{5-\sqrt5}{2}}}$$ $$-1+\sqrt5+\sqrt{5-2\sqrt5}$$ 27 3π/20
66 11π/30 $$\frac{1}{8} \left[\sqrt{30-6\sqrt5}+\sqrt5+1\right]$$ $$\frac12\left[\sqrt{10-2\sqrt5}+\sqrt3(-1+\sqrt5)\right]$$ 24 2π/15
67.5 3π/8 $$\frac{1}{2} \sqrt{2 + \sqrt{2}}$$ $$\sqrt{2} + 1$$ 22.5 π/8
69 23π/60 $$\frac{1}{16}\left[2(\sqrt3-1)\sqrt{5-\sqrt5}+\sqrt2(\sqrt3+1)(1+\sqrt5)\right]$$ $$\frac{1}{4}\left[2-(2-\sqrt3)(3-\sqrt5)\right]\left[2+\sqrt{2(5+\sqrt5)}\right]$$ 21 7π/60
72 2π/5 $$\frac{\sqrt{2+\phi}}{2}=\frac{\sqrt{5+\sqrt5}}{2\sqrt2}$$ $$\sqrt{5+2\sqrt5}$$ 18 π/10
75 5π/12 $$\frac{\sqrt{2+\sqrt{3}}}{2}$$ $$2+\sqrt{3}$$ 15 π/12
78 13π/30 $$\frac{1}{8} \left[\sqrt{30+6\sqrt5}+\sqrt5-1\right]$$ $$\frac12\left[\sqrt{10+2\sqrt5}+\sqrt3(1+\sqrt5)\right]$$ 12 π/15
81 9π/20 $$\frac12\sqrt{2+\sqrt{2+\phi}}=\frac12\sqrt{2+\sqrt{\frac{5+\sqrt5}{2}}}$$ $$1+\sqrt5+\sqrt{5+2\sqrt5}$$ 9 π/20
84 7π/15 $$\frac{1}{8} \left[\sqrt{10-2\sqrt5}+\sqrt3(1+\sqrt5)\right]$$ $$\frac12\left[\sqrt3(3+\sqrt5)+\sqrt{2(25+11\sqrt5)}\right]$$ 6 π/30
87 29π/30 $$\frac{1}{16}\left[(2+2\sqrt{3})\sqrt{5+\sqrt{5}}+\sqrt{2}(-1+\sqrt{5})(-1+\sqrt{3})\right]$$ $$\frac14\left[(2+\sqrt{3})(3+\sqrt{5})-2\right]\left[2+\sqrt{10-2\sqrt{5}}\right]$$ 3 π/60
90 π/2 1 $$\infty$$ 0 0
deg rad sin(a) = cos(b) = sin(90-b) = cos(90-a) tan(a) = cot(b) = tan(90-b) = cot(90-a) deg rad
a b

\begin{array}{r|l|l|l}
& deg
& \sin
& \cos
& \tan
\\
\hline 2\pi
& 360
& 0
& 1
& 0
\\
\hline \pi
& 180
& 0
& -1
& 0
\\
\hline \frac{2\pi}{3}
& 120
& \frac{1}{2}\sqrt{3}
& -\frac{1}{2}
& -\sqrt{3}
\\
\hline \frac{\pi}{2}
& 90
& 1
& 0
& \pm\infty
\\
\hline \frac{2\pi}{5}
& 72
& \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)
& \frac{1}{4}\left(\sqrt{5}-1\right)
& \sqrt{5+2\sqrt{5}}
\\
\hline \frac{\pi}{3}
&60
& \frac{1}{2}\sqrt{3}
& \frac{1}{2}
& \sqrt{3}
\\
\hline \frac{\pi}{4}
&45
& \frac{1}{2}\sqrt{2}
& \frac{1}{2}\sqrt{2}
& 1
\\
\hline \frac{2\pi}{9}
& 40
& \frac{i}{2}\left(\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}-\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}\right)
& \frac{1}{2}\left(\sqrt[3]{\frac{-1+\sqrt{-3}}{2}}+\sqrt[3]{\frac{-1-\sqrt{-3}}{2}}\right)
&
\\
\hline \frac{\pi}{5}
& 36
& \frac{1}{4}\left(\sqrt{10-2\sqrt{5}}\right)
& \frac{1}{4}\left(\sqrt{5}+1\right)
& \sqrt{5-2\sqrt{5}}
\\
\hline \frac{\pi}{6}
& 30
& \frac{1}{2}
& \frac{1}{2}\sqrt{3}
& \frac{1}{3}\sqrt{3}
\\
\hline \frac{\pi}{7}
&
& \frac{1}{24}\sqrt{3\left(112-\sqrt[3]{14336+\sqrt{-5549064193}}-\sqrt[3]{14336-\sqrt{-5549064193}}\right)}
& \frac{1}{24}\sqrt{3\left(80+\sqrt[3]{14336+\sqrt{-5549064193}}+\sqrt[3]{14336-\sqrt{-5549064193}}\right)}
&
\\
\hline \frac{2\pi}{15}
& 24
& \frac{1}{8}\left(\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}\right)
& \frac{1}{8}\left(1+\sqrt{5}+\sqrt{30-6\sqrt{5}}\right)
& \frac{1}{2}\left(-3\sqrt{3}-\sqrt{15}+\sqrt{50+22\sqrt{5}}\right)
\\
\hline \frac{\pi}{8}
& 22.5
& \frac{1}{2}\left(\sqrt{2-\sqrt{2}}\right)
& \frac{1}{2}\left(\sqrt{2+\sqrt{2}}\right)
& \sqrt{2}-1
\\
\hline \frac{\pi}{9}
& 20
& \frac{i}{4}\left(\sqrt[3]{4-4\sqrt{-3}}-\sqrt[3]{4+4\sqrt{-3}}\right)
& \frac{1}{4}\left(\sqrt[3]{4+4\sqrt{-3}}+\sqrt[3]{4-4\sqrt{-3}}\right)
&
\\
\hline \frac{\pi}{10}
& 18
& \frac{1}{4}\left(\sqrt{5}-1\right)
& \frac{1}{4}\left(\sqrt{10+2\sqrt{5}}\right)
& \frac{1}{5}\left(\sqrt{25-10\sqrt{5}}\right)
\\
\hline \frac{\pi}{12}
& 15
& \frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)
& \frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)
& 2-\sqrt{3}
\\
\hline \frac{\pi}{15}
& 12
& \frac{1}{8}\left[\sqrt{10+2\sqrt5}-\sqrt3(-1+\sqrt5)\right]
& \frac{1}{8} \left[\sqrt{30+6\sqrt5}+\sqrt5-1\right]
&
\\
\hline \frac{\pi}{18}
&10
&
&
&
\\
\hline \frac{\pi}{30}
& 6
& \frac{1}{8} \left[\sqrt{30-6\sqrt5}-\sqrt5-1\right]
& \frac{1}{8} \left[\sqrt{10-2\sqrt5}+\sqrt3(1+\sqrt5)\right]
&
\\
\hline \frac{\pi}{40}
& 4.5
& \frac{1}{2} \sqrt{2-\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}
& \frac{1}{2} \sqrt{2+\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}}
&
\\
\hline \frac{\pi}{45}
& 4
&
&
&
\\
\hline \frac{\pi}{60}
&3
& \frac{1}{16}\left[(2-2\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(-1+\sqrt{5})(1+\sqrt3)\right]
& \frac{1}{16}\left[(2+2\sqrt{3})\sqrt{5+\sqrt{5}}+\sqrt{2}(-1+\sqrt{5})(-1+\sqrt{3})\right]
&
\\
\hline \frac{\pi}{90}
&2
&
&
&
\\
\hline \frac{\pi}{180}
&1
& \frac1{2i}\left[\sqrt[3]{\cos(3^\circ) + i\sin(3^\circ)} – \sqrt[3]{cos(3^\circ) – i\sin(3^\circ)} \right]
&
&
\end{array}

The values for angles outside the range 0-90 degrees, can be found by reducing the angle be in the correct range. Negative angles should have 360 added to them until positive. If angle ≥360, subtract multiples of 360 from it, until <360. Then, if angle ≥270, subtract from 360. If angle ≥180, subtract 180. If angle ≥90, subtract from 180.

Examples:

• sin(-54)=sin(360-54)=sin(306)=-sin(54)
• cos(120)=cos(180-120)=-cos(60)
• tan(225)=tan(225-180)=tan(45)
• tan(-225)=tan(360-225)=tan(135)=tan(180-135)=-tan(45)
• sin(1000)=sin(1000-360*2)=sin(280)=sin(360-280)=-sin(80)

Angles between 0 and 360 will have the following signs and are measured in a counter-clockwise arc originating at the “x+” line.

FYI, here is the exact value of sin 1°, $$\frac1{2i}\left[\sqrt[3]{\cos(3) + i\sin(3)} – \sqrt[3]{cos(3) – i\sin(3)} \right]$$, where $$i=\sqrt{-1}$$,

$$\frac12(-1-i\sqrt3)\sqrt[3]{-\frac{\sqrt6}{384} (\sqrt5-1)(3+\sqrt3)+ \frac{\sqrt3}{192}(3-\sqrt3)\sqrt{5+\sqrt5}+\frac{i}{8}\sqrt{1-\frac1{48}\left[\sqrt6(\sqrt5-1)(3+\sqrt3)-2\sqrt3(3-\sqrt3)\sqrt{5+\sqrt5} \right]^2} } + \\ \frac12(-1+i\sqrt3)\sqrt[3]{-\frac{\sqrt6}{384} (\sqrt5-1)(3+\sqrt3)+ \frac{\sqrt3}{192}(3-\sqrt3)\sqrt{5+\sqrt5}-\frac{i}{8}\sqrt{1-\frac1{48}\left(\sqrt6(\sqrt5-1)(3+\sqrt3)-2\sqrt3(3-\sqrt3)\sqrt{5+\sqrt5} \right)^2} }$$

and the cos 1°,

$$\frac12\sqrt{2- (i\sqrt3-1)\sqrt[3]{\frac{-\sqrt3}{64}(1+\sqrt5)-\frac{\sqrt2}{128}(\sqrt5-1)\sqrt{5+\sqrt5} + \frac{i}{8}\sqrt{1-\left[\frac{\sqrt3}{8}(1+\sqrt5)+\frac{\sqrt2}{16}(\sqrt5-1)\sqrt{5+\sqrt5}\right]^2}} + \\(i\sqrt3+1)\sqrt[3]{\frac{-\sqrt3}{64}(1+\sqrt5)-\frac{\sqrt2}{128}(\sqrt5-1)\sqrt{5+\sqrt5} – \frac{i}{8}\sqrt{1-\left[\frac{\sqrt3}{8}(1+\sqrt5)+\frac{\sqrt2}{16}(\sqrt5-1)\sqrt{5+\sqrt5}\right]^2}} }$$