# How do you find exact values for the sine of all angles?

By Murray Bourne, 23 Jun 2011

**Challenge: **What is the exact value for sine of 6 degrees? How about sine of 1 degree?

**Context: **I received a delightful email from reader James Parent recently. He wrote:

I have the exact answers for the sin of all integer angles. Has anyone done this before? I’m retired, and a Professor Emeritus from a community college. I’m 74 years-old.

This certainly sounded interesting to me, so I asked James to write a guest post, and here it is. (Many of James’ mails had the tag-line “Sent from my iPad”.)

Over to James.

## How do you find exact values for the sine of integer angles?

Here is one way of going about it.

**Background**

Let’s find some **exact values** using some well-known triangles. Then we’ll use these exact values to answer the above challenges.

**sin 45****°: **You may recall that an isosceles right triangle with sides of 1 and with hypotenuse of square root of 2 will give you the sine of 45 degrees as half the square root of 2.

**sin 30****° and sin 60****°: **An equilateral triangle has all angles measuring 60 degrees and all three sides are equal. For convenience, we choose each side to be length 2. When you bisect an angle, you get 30 degrees and the side opposite is 1/2 of 2, which gives you 1. Using that right triangle, you get exact answers for sine of 30°, and sin 60° which are 1/2 and the square root of 3 over 2 respectively.

## Using these results – sine 15°

How do you find the value of the sine of 15°?

Sine of half an angle in the first quadrant is given by the expression:

So the sine of 1/2 of 30° will be:

which gives us

or

**Note: **We could also find the sine of 15 degrees using sine (45° − 30°).

**sin 75****°**: Now using the formula for the sine of the sum of 2 angles,

sin(*A* + *B*) = sin *A* cos* B* + cos *A* sin *B*,

we can find the sine of (45° + 30°) to give sine of 75 degrees.

We now find the sine of 36°, by first finding the cos of 36°.

**cos 36****°: **The cosine of 36 degrees can be calculated by using a pentagon. See

http://www.cut-the-knot.org/pythagoras/cos36.shtml where it is shown that

Putting these values on a right triangle and solving for the unknown side, we can conclude:

**sin 18****°: **Now, the sine of 18 degrees comes from the sine of half of 36 degrees.

Calculating this, the sine of 18 degrees becomes

**sin 3****°: **The above leads you to one of the paths to sine of 3 degrees and to sine of 6 degrees.

For example, sine (18° - 15°) will give us the sine of 3 degrees. which is

sin 3° = sin (18° − 15°) = sin 18° cos 15° − sin 15° cos 18°

This gives us the following value of sin 3°:

or other forms depending how you factor the above.

**sin 6****°****: **Using the above, one can compute the sine of 6 degrees finally as sine of twice 3 degrees to arrive at

**sin 18****° and sin 72****°:** Taking the equivalent sine and cosine values of 15° and 18° on the right hand side of

sin 3° = sin (18° − 15°) = sin 18° cos 15° − sin 15° cos 18°

gives us:

sin 3° = sin 18° sin 75° − sin 15° sin 72°

We can calculate the values of the sines of 18° and 72° from the above expression.

**Sines of other angles**

Many angles can be computed exactly by many methods. Another practical formula is the sine of 3 times an angle:

sin 3*A* = 3 sin *A* − 4 sin^{3}*A*

**sin 9****°: **For example, the sine of 9 degrees is the sine of (3×3°).

So, with A = 3, we arrive at

And so on.

**sin 1****°: **Now, to find the sine of one degree, one needs to know sine of one third of three degrees!

One needs to solve the above for sin (A) in terms of 3A, and this involves solving the cubic. As you know, the cubic was solved many, many years ago.

There are three solutions and one needs to know which one to use and when! Experience has taught me to use the following for a quadrant I angle (the "*I*" in this expression stands for the imaginary number √(−1). See Complex Numbers for more information.)

[Click image to see full size]

Use the following when you have a quadrant II angle:

Use the following for quadrant III angles:

[Click image to see full size]

So, the expression for sine(1°) becomes

[Click image to see full size]

Messy, isn’t it! But, it does give you the exact value for the sine of one degree.

## Is it correct?

Evaluate the sine of 1 degree using a TI Scientific Calculator and you will get 0.0174524064. Evaluate the above messy expression and you will also get 0.0174524064. Even allowing for calculator rounding errors, we can be confident our answer is correct.

## List of all sines of integer degrees from 1° to 90°

This PDF contains all the exact values of the sine values for whole-numbered angles (in degrees):

Exact values sin 1° to sin 90° [PDF, 293 kB]

**Concluding Comments from James **

For a retired community college mathematics professor since 1997, this has been a lot of enjoyment for me.

James Parent, Professor Emeritus

Schenectady County Community College, Schenectady, New York

Currently teaching as an adjunct at Great Bay Community College, Portsmouth, New Hampshire

See the 45 Comments below.

24 Jun 2011 at 5:56 pm [Comment permalink]

Unbelievable.

25 Jun 2011 at 1:02 am [Comment permalink]

An easy method to memorize the sequence sin(0°), sin(30°), sin(45°), sin(60°), sin(90°). Here is it:

1. Write below five division lines the digit 2:

2. Write the digits 0, 1, 2, 3, 4 at the top

3. Sqrt the top digits:

4. You are ready:

sin(0°) = 0

sin(30°) = 1/2

sin(45°) = √2/2

sin(60°) = √3/2

sin(90°) = 1

25 Jun 2011 at 9:48 am [Comment permalink]

Thanks, Philip! This one is new to me. We all have different ways of learning and understanding things.

I never try to memorize these exact values - I just sketch the diagrams quickly and read them from there.

30 Jun 2011 at 12:46 am [Comment permalink]

This is a nice piece of work and possibly quite engaging to some of our high school students taking Trig. My only critique is kind of nitpicking, but the title refers to “any” angle, while the professor clarifies later to any “integer” angle (in degrees). Until he cleared it up, my first thought was “What about irrational angles?

1 Jul 2011 at 3:01 am [Comment permalink]

This is wonderful! Easy remembrance of trig ratio of special angles INDEED !

2 Jul 2011 at 6:23 am [Comment permalink]

@Douglas: The title was already quite long, that’s why the word “integral”was not included. Also, it made the piece look less palatable!

2 Jul 2011 at 1:25 pm [Comment permalink]

“integral” = “integer” ?

10 Jul 2011 at 4:26 pm [Comment permalink]

Certainly amused by the amount of detail it goes to. I hate remembering stuffs, and taking the extra mile to understand is totally worth it. I’m having a problem understanding how this formula works though.

Ok I know why the value of cos 30 is the square root of 3 over 2, but why the other stuffs around it?

11 Jul 2011 at 8:26 am [Comment permalink]

@Bonzoni: You can find an explanation for this and other “half angle” formulas here: Half angle formulas.

27 Jul 2011 at 7:01 am [Comment permalink]

Wow this is an amazing find. This list is certainly powerful using it one can easily calculate the cosine values because sine and cosine are cofunctions. Next tangent can be calculated via tan(x) = sin(x)/cos(x). Then getting the reciprocals is easy work. For any integral angle other than 1 through 90° you simply apply the quadrant rules. Very cool work.

3 Apr 2012 at 11:40 am [Comment permalink]

How do you solve cos36 degree without using a calculator ?

3 Apr 2012 at 3:34 pm [Comment permalink]

@Keng Keh. The answer is in the article, about 1/2 way down. There is a link which shows how to do it via a pentagon.

5 Apr 2012 at 9:46 am [Comment permalink]

Thank you Mr Murray, I got it !

27 Apr 2012 at 10:29 am [Comment permalink]

Thanks a lot !!!

Onto to my new Maple 16 and Mathematica 8.0.4 to calculate the values for cosine.

18 Jun 2012 at 5:00 pm [Comment permalink]

Why the value of sin30 is half?

19 Jun 2012 at 8:43 am [Comment permalink]

@Raj: This page has the answer

Values of Trig Functions

20 Jun 2012 at 3:07 am [Comment permalink]

how will give the value in bisection method cos1 value ?

9 Aug 2012 at 10:37 am [Comment permalink]

Thank very much for the work done by Professor James Parent, I note that there is one typing mistake for sin3A which should be equal to 3sinA - 4(sinA)^3

9 Aug 2012 at 10:49 am [Comment permalink]

@Hans - Thank you for spotting the error. I have amended the article.

21 Sep 2012 at 2:13 pm [Comment permalink]

Sorry to take up so much comment space. Please reply if you can add to this observation.

Mr. Bourne and Prof. Parent;

I worked for many years as a video shop and field technician and used complex waveforms on oscilloscopes to align machines, which drew somewhat on my math background from high school and early college. We used sine waves and lissajous figures, which were rather simplified on the graphic screen of the scopes. We didn’t do that much calculation, as it were.

Much of that work has dried up and now I’ve gone back to brush up on my math skills and perhaps may even go back to school. While trying to regain my algebra, trigonometry, and calculus I started working on the unit circle, and working out the sine and cosine and tangent of the basic angles. I looked around on the internet for examples and found some great examples and interesting reading on your math page. I have something that I hope may be a worthwhile contribution to the pages.

I was intrigued by the project on your page where a Professor Emeritus James Parent showed how to derive the constant values for the sines of all integer angles, using various formulas and methods. I worked for a while to understand how he got 36 and 18. Then I went back and looked at 15 and 75, which are a bit easier to understand. While looking at his derivation of sine of angle 15 I noticed that he came up with an answer that didn’t fit my calculation using the half angle rule.

In Professor Parent’s commentary, he works out sine 15 using the half angle formula sq root ((1 - cos 30) / 2), and shows that it simplifies to 1/2 x sq root( 2 - (sq root 3)), which he says, quite nonchalantly “or”, further simplifying the expression, to 1/4 x sq root 2 x (sq root 3 - 1).

The simplification, using algebra, to go from the first expression to the second, is quite tricky. I think he may have put in the second expression without any fanfare, since both quantities are used for the value of sine 15.

It’s interesting that his solution to the problem of sin 15 leads to a way to denest the radical in the expression sq root (2 - (sq root 3)). Without going through the process of squaring, expanding, and using the quadratic formula as I’ve seen performed, you can use a kind of trigonometric substitution to get back to sin 15. Then you can solve for sin 15 with the difference formula, writing sin 15 = sin (45 - 30). Using the formula sin45 x cos 30 - cos45 x sin 30, you will come up with sq root (6) / 4 - sq root (2) / 4, which can be written as 1/4 x sq root 2 x (sq root 3 - 1), the Professor’s end answer for sine 15.

If one were to start with a simple, commonly seen nested radical of sq root (2 - ( sq root 3)), you could multiply by 1/2, or 1/sq root 4, and that would get you to sq root ((2 - sq root 3) / 4). This can be rewritten as sq root ((1 - cos 30) / 2). This equals sin 15 by the half angle theorem. Now substitute sin 15 by the difference theorem, using sin (45 - 30). This will produce the denested answer 1/2 x sq root( 2 - (sq root 3)).

A similar trigonometric substitution can be used to denest (2 + ( sq root 3)), using cos 15, translating from the half angle formula of cos 30 and the difference formula of cos 45 - 30.

It works for angles 150 and 75, although you get the same answer values. I don’t see any other substitution denest shortcuts very readily. It would be anywhere you could get an angle by the half angle formula, which often produces a nested radical, and get the same angle by a sum or difference formula. I’ll look for others in the degrees and pi fractions.

Perhaps a similar manipulation can be done between the three times or one third formula, and a sum or difference of angles that have a product and sum in common.

What do you think of this method? Is this a known or often used substitution to simplify a nested radical? It seems no less magical slight of hand than going through the quadratic formula.

If you need to see a more completely written out form of the math I can send. It would be nice to do it in a format that has the radicals in a printed form so it’s more clear.

-Keith J.

21 Sep 2012 at 8:44 pm [Comment permalink]

Hi Keith

Thanks for your inputs! Yes, some of the formulas could be simplified. Of course, at the end of the day the main point was to show that it can be done.

20 Nov 2012 at 3:51 pm [Comment permalink]

I’m impressed by your compilation of exact sine table. However, in the mission to derive the exact algebraic expression of sin(1°), we reach a difficult destination: a complicated form of radicals which contains the IOTA. If we try to remove it in order to get a cleaner IOTAless form, we have to apply deMoivre or binomial expansion, both of which lead to a form which is the sum of a sine series and a cosine series. Conclusion is that, we cannot achieve an algebraic expression of sin(1°) which contains only surds and rational numbers as we have for angles which are multiples of 3°.

My question is that, in such a situation, how can we say that sin(1°) is irrational when it appears to be a transcendental number? Just because 1° is a rational multiple of pi, should sin(1°) be declared irrational instead of transcendental?

27 Nov 2012 at 8:30 am [Comment permalink]

@Pinaki ROY 20 Nov 2012 at 3:51 pm,

See if this helps:

I remember vaguely that for nonzero x, both x and sin x cannot be algebraic. It might have been in S. Lang’s Transcendental Numbers. (Here x of course is in radians). Irrational numbers are partitioned into algebraic and transcendental.

18 Dec 2012 at 11:54 pm [Comment permalink]

In fact, it is not the sin of 36, rather the cosine of 36 which is (1 + sqrt(5))/4

19 Dec 2012 at 9:02 am [Comment permalink]

Good catch, Marshall! Hmmm - If anyone has plenty of time, please let me know of any other problems with all these calculations and I’ll tidy it all up in one go.

28 Dec 2012 at 1:31 am [Comment permalink]

Glad I could help… I teach at a university, mostly calculus 1 and 2……. I make up a very detailed key to my tests, including a detailed grading standard for awarding points….. never taking them away….. but the relevant remark is this: I think the last time I got a 100 on one of my own tests was sometime before the Civil War….. hehehe….keep up the good work!!!!

23 Jan 2013 at 12:27 pm [Comment permalink]

I am not certain the answer for sin(1) (I mean one degree) is correct. Your answer is of the form

where

This simplifies to

where

When you compute the cubic root of a number, you get three possibly complex answers. Therefore, your answer will result in the following:

with k=0,1,2.

There are two problems with this. First, none of them is cos(89) = sin(1). Second, even if you had cos(89) as one of the answers, all you would have shown is that sin(1)=cos(89)=sin(1), which is trivially true.

I believe the problem for all answers you have provided for angles that are not multiple of 3 degrees is that you have cubic roots of complex numbers in them. Those cubic roots will end up to be of the form re**(j*phi), and you’ll end up showing sin(x) = sin(x).

Am i missing something here?

8 Feb 2013 at 3:03 pm [Comment permalink]

[…] Determining sine for all angles […]

14 Apr 2013 at 1:36 am [Comment permalink]

This is a great list of exact values. Professor James Parent is a regular Ramanujan.

23 Apr 2013 at 4:51 pm [Comment permalink]

the answer equations in the form as provided by Nader Moayeri in 23.1.2013 are incorrect in both the answer equations and the expression for ‘a’, they are not the exact replica of the solution of sin(1 degree) as provided by James Parent. The solution by James making two complex roots adding togther may be misleading, the solution for cubic equation provided by T. Irvine in his website http://www.vibrationdata.com is good for reading.

25 Apr 2013 at 10:38 pm [Comment permalink]

Fantastic reading! I have a question about the second comment though - I too have used the method for “easy” exact values where you write the numbers 0-4, put each of them over 4 and take the square root, however I cannot find anywhere about where this comes from or why it works! Can anyone enlighten me? I don’t like teaching things when I am unsure of their origin, even when I know it works!

26 Apr 2013 at 6:10 am [Comment permalink]

@Siobhan: Well, the origin is the ratios for the 30-60 and 45-45 triangles as mentioned in the article. From there, it’s just an observation.

I think it’s the same kind of question as “Why does a 3-4-5 triangle work to give us a right angled triangle?” It’s just one observation of a specific case of Pythagoras’ Theorem.

28 May 2013 at 9:14 am [Comment permalink]

I thank Hans Chan (see his post of April 23, 2013) for pointing out errors in what I had posted on January 23, 2013. This is an attempt to fix the errors and set the record straight.

The first expression in my earlier post was correct, but the expression for a was incorrect. Here’s the correct version:

a = (1/24)(3)^(1/2)(3-(3)^(1/2))((5+(5)^(1/2))^(1/2))

- (1/48)(6)^(1/2)((5)^(1/2)-1)(3+(3)^(1/2)) = -0.052335956

and the correct value for theta is

theta = 93 degrees

When you compute the cubic root of e^(j theta) and e^(-j theta), each will have three cubic roots. This leads to nine possible combinations. Six of these will result in complex numbers as the answer for sin(1 degree), which is obviously unacceptable. Three will result in real numbers and turn out to be cos(89 degrees), cos(31 degrees), and cos(29 degrees). Obviously, the last two are wrong answers. The correct answer is sin(1 degree) = cos(89 degrees), but this is trivially true.

The problem comes from the fact that Professor Parent’s expression for sin(1), albeit mathematically correct, requires computation of cubic roots of two complex numbers. I do not know of any way of computing the n’th root of a nonzero complex number other than converting the complex number to the form rho e^(j theta), where rho > 0 and 0 <= theta < 2 pi, and then writing the roots (there will be n of them) as

(rho)^(1/n) e^(j(theta + 2k pi)/n); k= 0, 1, …, n-1

but the expression for theta includes an inverse tangent function. So, the only way I know how to compute the expression given by Professor Parent for sin(1) is to hit that tan^(-1) key on the calculator and then I would need to hit the cos key also, because the final answer will be of the form of cosine of some angle, 89 degrees to be specific.

If any of you know how to compute the cubic root of a complex number without using an inverse trigonometric function, please let the rest of us know. If such a procedure does not exist, then it would be reasonable to conclude that the expression given for sin(1) is just a mathematical curiosity, because it cannot be computed without invoking inverse trigonometric and trigonometric functions. The whole point of Professor Parent's work was to find expressions for sin of integrer-valued (in degrees) angles using the four basic arithmetic functions (+, -, *, and /) and square and possibly cubic roots of real numbers. Otherwise, to compute sin(1), all you have to do is hit the 1 button on your calculator followed by the sin button.

28 May 2013 at 9:39 am [Comment permalink]

Perhaps the following would be of interest to this group. I derived the following formulas. I am sure these must be known. SO, I suspect I have reinvented the wheel!

Anyway, this is about finding the square roots of the complex number a+jb. The answer is plus minus (I don’t know how to do this without causing confusion)

(((a^2+b^2)^(1/2)+a)/2)^(1/2) + j (((a^2+b^2)^(1/2)-a)/2)^(1/2)

if b >=0, and plus minus (because each complex number has two square roots)

(((a^2+b^2)^(1/2)+a)/2)^(1/2) - j (((a^2+b^2)^(1/2)-a)/2)^(1/2)

if b < 0.

The point is that you can find the square root of a complex number of the form a+jb directly and without the ned to use trigonometric and inverse trigonometric functions.

I tried hard to derive similar expressions for the cubic root of a+jb, but all my attempts failed. In the 2-3 approaches I pursued, eventually the trigonometric and inverse trigonometric functions crept in as a result of the need to solve a cubic equation.

Well, for what it's worth I decided to share the above with you, just in case you have not seen it.

29 May 2013 at 9:18 am [Comment permalink]

Nader Moayeri has pointed out the puzzle that cubic root of complex number may not generate the meaningful result as wanted. I thank for his work to find sensible answer. However complex number is a field out of real number, so its geometric interpretion is uneasy to be visualized.

1 Jun 2013 at 12:09 pm [Comment permalink]

I am afraid I do not understand what Hans Chan is saying in his May 29, 2013 post. This has nothing to do with one or more choices for the two cubic roots that have appeared in Professor Parent’s expression for sin(1) resulting in wrong answers. In many physical problems you solve an equation and you discard some of the roots because they are not acceptable. What matters is that one set of choices for the two cubic roots results in a value of cos(89) = sin(1) for the expression provided by Professor Parent, and that is obviously the right answer.

This has nothing to do with geometric interpretation of complex numbers either. We are really not looking for a geometric interpretation.

What matters is that Professor Parent’s goal was to derive an expression for sin(1) whose computation requires only the four basic arithmetic operations and square roots (see the expressions he has provided for sin(45), sin(30), sin(60), sin(75), sin(15), sin(36), sin(18), sin(3), sin(6), and sin(72)). However, I maintain that the computation of the expression he has provided for sin(1) involves using an inverse trigonometric function followed by a sin or cos (trigonometric functions). So, Professor Parent’s expression for sin(1) is of a different nature than the expressions he has provided for all those other angles just listed. It does not meet the requirement of needing only +, -, *, /, and square roots. If someone can find an expression for sin(1) that meets that condition, then I would say he/she has solved the problem. (p.s. A Taylor series expansion is not acceptable!) Until then, the problem remains unsolved.

24 Oct 2014 at 11:04 am [Comment permalink]

Dear James Parent: This was extremely helpful to me! Thank you so much for taking the time to post it. LLAP

1 Nov 2014 at 2:18 am [Comment permalink]

Sine 36 value is WRONG

Actually COS 36 is what you put in place of SINE 36

1 Nov 2014 at 2:30 am [Comment permalink]

Hi All,

Sorry for being so abrupt in my previous comment.

I went through derivation given in the link and when I came back sin 36 seemed to have a value for cos 36.

cos 36° = (√5+1)/4

Sin 36 = sqrt(2(5−√5))*(1/4)

sin 18 = (−1+√5)/4

So please look into it.

Regards

RJ

18 Nov 2014 at 9:34 am [Comment permalink]

@Jagadeeswaran: You were right that the sine 36° value was incorrect and so I have amended the post. (Your value for sine 36° had reversed 5 and √5, which I also fixed.)

Thanks for your inputs!

19 Nov 2014 at 6:01 pm [Comment permalink]

whats the sin6.27’?

20 Nov 2014 at 8:53 am [Comment permalink]

Do you mean the exact value of sin 6.27? This is quite a challenge - anyone like to have ago?

27 Nov 2014 at 10:33 pm [Comment permalink]

sin6°27′

Let’s forget the 6° and think about 27′ = (27/60)°

= (1/5)(3/4)*3°

The 1/5 means that we have to solve a quintic equation:

sin5x = 5sinx - 20sin^3x + 16sin^5x (where sin5x is known)

And that can’t be done, as was shown already by Abel.

14 Mar 2015 at 7:22 am [Comment permalink]

Very interesting, thanks for sharing. Have you ever considered calculating these in terms of phi? It may produce more aesthetic solutions, and lead to general conclusions. We may be able to have mathematica (or similar) do all the work, but if not it’s pretty fun to algebraically manipulate phi. If it hasn’t been done, I’d be happy to work on it with you, if you’re interested.

15 Mar 2015 at 4:02 pm [Comment permalink]

@Brian: Thank you for your offer. I’d be very happy if you would like to go ahead. I agree that getting Mathematica’s help would be the way to go.

As mentioned a bit earlier, I know there are some errors in the current document, but I unfortunately don’t have the time at the moment to dig through and correct them. If you notice any of them as you go, your suggestions for corrections would be appreciated!