Exact Trigonometric Function Values

A Table of Exact Trig values

All the trig functions in one diagram

Mathematica

However, the diagram is useful for seeing what is measured by the six functions but it does not give any indication of the signs of the values.
The signs are chosen to make the trig formula see below consistent for all angles. Also, the next section shows their graphs and from the basic SINE graph, all the others follow.
The sine function has many applications in mechanics (e.g. the motions of rotating objects), electronics (e.g. alternating electric current and electromagnetic waves), ... .

Trig functions of Angles outside the range 0° to 90°

The basic formulae here all reply on the SINE graph as follows:

• The sine of an angle is defined by the vertical height of a point as it rotates around a unit circle (that is, its radius is 1) measured from a horizontal line through the centre of the circle. So it cannot be bigger than 1 or less than -1.
• the cosine of an angle is defined by the horizontal distance of a point as it rotates around the unit circle measured from a vertical line through the centre of the circle. It too must be in the range -1 to 1.
• From the diagram it is easy to see that the sine and cosine graphs are the same shape, but displaced by a quarter turn (90°). In mathematical terms, we would say cosine(x) = sine(90°+x).
• The tangent of an angle measures the ratio of sine to cosine and so can take any value whatsoever.
  At multiples of 90° it becomes infinite or negatively infinite.

The angles can be measured by several types of unit.

• Often this is degrees with a complete turn divided into 360 degrees
• If we define sine and cosine by distances (or coordinates) of a point on a unit circle, we can also define the angle by a distance on that circle too: the distance on the circumference that a point travels in turing through that angle. The circumference of a unit circle is 2 π. This measure is called a radian.

Patterns

The Simple Square-Root pattern

The √2 ± √n pattern

The √2 ± Phi pattern

The Golden Ratio

A more general pattern

Proofs

30° 45° and 60°

This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.	Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.

36° and 54°, 18° and 72°

The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.   The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.

15° and 75°

	a2	= b2 + c2 – 2 b c cos(A)
	For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:
	a2	= 22 + 22 – 2 x 2 x 2 x cos(30°)
		= 8 – 4 √3
		= 2 (4 – 2 √3)
	But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and so
	a2	= 2 (√3 – 1)2
	Taking the square-root:
	a	= √2 (√3 – 1) which we can also write as
		= 2 (√3 – 1) / √2
	Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:

Here is another diagram with its lengths calculated easily. From it we can read that

• tan(15°) is 1/(2+√3) which is also equal to (2 − √3)
• tan(75°) is (2+√3)

Reference

• Note 108.10 Proof without words, tan(&pi/12) = 2 - √3, tan(5π/12) = 2+√3 M Lukarevski, Math Gaz vol 108, (571) March 2024, pages 143-144.

Ailles Rectangle for 15° and 75°

Show times √2

Trig Formulae

`sin(x) = a / h ` ` cos(x) = b / h` `"cosec"(x) = h/a` ` sec(x)=h/b` `tan(x) = a / b` `cot(x) = b / a`

`\color{blue}{sin(x) = cos(90° - x)}`   `\color{blue}{sin^2(x) + cos^2(x) = 1}` `\color{blue}{tan(x) = sin(x)/cos(x)}`   `\color{blue}{cot(x) = 1/tan(x) = cos(x)/sin(x)}` `\color{blue}{tan^2(x) + 1 = 1/(cos^2(x))}`   `\color{blue}{cot^2(x) + 1 = 1/(sin^2(x))}`

If we know the value of a trig function on two angles `a` and `b`, we can determine the trig function values of their sum and difference using the following identities:

If the two angles are the same (i.e. `a=b`) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:

A diagram to relate many angles and Phi

Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle.
Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).

Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next page.

You do the maths...

Other angles with exact trig expressions involving square-roots

Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.

Halving

The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:

`\color{blue}{cos(pi/4)}` = `\color{blue}{(sqrt 2)/2}`
`\color{blue}{cos(pi/8)}` = `\color{blue}{(sqrt (2+sqrt 2))/2}`
`\color{blue}{cos(pi/16)}` = `\color{blue}{(sqrt(2+sqrt (2+sqrt 2)))/2}`
`\color{blue}{cos(pi/32)}` = `\color{blue}{(sqrt(2+sqrt(2+sqrt (2+sqrt 2))))/2}`

Superimposing

Do we know all the angles?

In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.

Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its prime factors to solve two equivalent problems:

• drawing a regular n-sided polygon using only a straight edge and compass and
• expressing the cos and sin of 360/n° using only square roots.

• b,c,... and all the powers except a , the power of 2, must be 1, and
• the primes>2 that are factors of n (that is p₁, p₂, ...) must be of the form 2^2k+1 for some number k.

Prime numbers of the form 2^2k+1 are called Fermat primes. The series of numbers of the form 2^2k+1 begins

Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:

The odd terms (the left hand column apart from 2) is the series

Some interesting facts about these odd numbers (we include 1 here too) are:

• all these 31 odd numbers are divisors of 2²⁵–1 = 2³² – 1 = 4294967295 (see A004729)

1 = 12 3 = 112 5 = 1012 15 = 11112 17 = 100012 51 = 1100112 85 = 10101012 255 = 111111112	1  1  1  1  2  1  1  3  3  1  1  4  6  4  1  1  5 10 10  5  1  1  6 15 20 15  6  1  1  7 21 35 35 21  7  1
	the odd numbers	8 16 32 64 128 rows

The π/17 formula

The next value with an odd denominator is π/257 and the exact expression for cos(π/257) using square-roots only will be even more complicated!

Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book (1988).

The π/15 formula

A Trig Calculator using Decimal numbers of any length

For other values, such as `\color{blue}{cos(pi/8)}`, type in and evaluate it first as a decimal and then as a continued fraction showing that the CF is [0; 1, 12, 7, 3, 2, 1, 1, 1, 1, 20, 5, 3, 11, 1, 7, 7, 3,...]. Even looking ahead for many more terms there is no pattern here.
However, if we change the Expression to Evaluate to the CF this time is [0; 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4,...].
There is a clear pattern here that goes on for many terms. If we delete the 1,4,1,4,... part and put it into the recurring part input for the CF so that you have the CF [0; 1, 5, 1, 4] and then pressing the convert-to-fraction-form button ( ) will show `\color{blue}{(2+sqrt(2))/4}` as the value of `cos(pi/8)^2` and so we deduce that (computationally) `\color{blue}{cos(pi/8)=sqrt(2+sqrt(2))/2}`.

In this way we can experimentally discover many such expressions for trig forms using the ability of this Calculator to quickly evaluate to many decimal places and to (accurately and exactly) convert continued fractions to the form `(a\pm sqrt(b))/c`. Then we can look for a mathematical proof that our results are indeed true.
Change the Working Precision to 50, 100, or even 200 dps to see if patterns persist with more computational accuracy.

References

• 66.41 Some Trigonometric Ratios in Surd Form J M H Peters The Mathematical Gazette Vol. 66, (1982), pages 296-299
  though there are some errors in his formulae involving τ (our Phi) which are corrected in the equations on this page.