# Rational with finite decimals values for sine, cosine, and tangent

What are the possible combinations of sine, cosine, and tangent values such that all three are simultaneously rational with finite decimals?

I am aware of the below two cases.

• $$\sin(x) = 0, \cos(x) = 1, \tan(x) = 0$$ is a trivial case for angle measure zero

• $$\sin(x) = 0.6, \cos(x) = 0.8, \tan(x) = 0.75$$ is another case, for an angle of $$37$$ degrees (approximately)

Are there any other combinations that satisfy the condition? If not, is it possible to prove that these are the only possible combinations?

Edit: Sorry, I missed one detail when posting the question. I am looking for sin, cos and tan values being rational numbers with finite decimals. For example, (7, 24, 25) is a Pythagorean triple that does not satisfy the condition because tan value would be 0.291666....

• Any Pythagorean triple will give such values. – Blue Feb 20 '19 at 6:55
• With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others? – Henry Feb 20 '19 at 8:36
• Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal). – PM 2Ring Feb 20 '19 at 8:58
• Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none – Henry Feb 20 '19 at 10:13
• @PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10. – Acccumulation Feb 20 '19 at 16:25

Any Pythagorean triple will give you rational $$\sin, \cos, \tan$$ values, and any triple of rational trig values will give you a Pythagorean triple.

The Pythagorean triples have been completely classified. Take two natural numbers $$u>v$$. Then $$(u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2$$ and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $$\sin$$ and $$\cos$$ pairs are all possible values of $$\frac{u^2-v^2}{u^2+v^2}\text{ and }\frac{2uv}{u^2+v^2}$$

• You are right, But I missed a detail in my question - it is now edited. – Shahul Feb 20 '19 at 8:00
• @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$. – Arthur Feb 20 '19 at 8:02
• In order to guarantee $\tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively. – Wojowu Feb 20 '19 at 9:20
• @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $k\geq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$. – Servaes Feb 20 '19 at 9:37
• I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution? – user1008646 Feb 21 '19 at 6:20

A hypotenuse-$$1$$ right-angled triangle containing an angle $$x$$ has side length $$\sin x$$ opposite it and $$\cos x$$ adjacent to it, and if these are rational so is their ratio $$\tan x$$. But these lengths are rational iff the triangle can be scaled to have integer side lengths, so the rational choices for $$\sin x,\,\cos x,\,\tan x$$ with $$x$$ acute are precisely those obtained from Pythagorean triples. Extending to arbitrary angles so some functions may be negative, the general solution is $$\sin x=\frac{2ab}{a^2+b^2},\,\cos x=\frac{a^2-b^2}{a^2+b^2},\,\tan x=\frac{2ab}{a^2-b^2}$$ with $$a,\,b\in\mathbb{Z},\,a\ne\pm b$$.

Note that

$$\frac{\sin(x)}{\cos(x)} = \tan(x)$$

Therefore, whenever sine and cosine are rational, and cosine nonzero, tangent is also rational. Since sine and cosine are also continuous for all real numbers $$x$$ and are what we call "onto" or "surjective" on the interval $$[-1,1]$$, for each number in that interval we can find $$x$$ such that $$\sin(x)$$ and $$\cos(x)$$ are that number.

We also know that the rationals are "dense" in the reals - that is, in between any two real numbers, you can always find a rational number. It ergo goes that we can find infinitely many $$x$$ combos so that $$\sin(x)$$ and $$\cos(x)$$ (independent of each other) are rational.

This doesn't really address the case of their ratio or them being rational at the same time, though.

For this, as noted by Blue in the comments, Pythagorean triples will always work. Pythagorean triples define three sides of a right triangle, specifically positive integer solutions to $$a^2 + b^2 = c^2$$. Because the triangle is a right triangle, we can also note that any $$x$$ we use as the angle here (opposite a leg) will be between $$0$$ and $$\pi/2$$ in measure, noninclusive.

There is a method to generate infinitely many Pythagorean triples too! Take positive, unequal integers $$m,n$$ - any two you want. Then $$a = 2mn, b = m^2 - n^2, c = m^2 + n^2$$. Since $$m,n$$ are integers, then $$a,b,c$$ are integers. Then in turn $$\sin(x) = b/c$$ (or $$a/c$$, depending on the choice of angle), and $$\cos(x) = a/c$$ (or $$b/c$$), giving those two are rational. From that, we see that $$\tan(x) = b/a$$ (or $$a/b$$) and thus is also ratio.

Thus, for each pair of unequal, positive integers $$m,n$$ we can have

$$\sin(x) = \frac{b}{c} = \frac{m^2 - n^2}{m^2 + n^2} \;\;\; \cos(x) = \frac{2mn}{m^2 + n^2} \;\;\; \tan(x) = \frac{m^2 - n^2}{2mn}$$

giving infinitely many solutions such that $$\sin(x),\cos(x),\tan(x)$$ are all rational. From there, finding the corresponding $$x$$ is trivial.

As for whether these are the only solutions, I do not know.

Have you tried using base $$6$$? I have heard that doing so can result in more accurate results when using $$\pi$$. It may also work better in this case, as the desired result are finite decimals.

• Welcome to Math.SE. How is this relevant to the question. Please elaborate – Shailesh Feb 21 '19 at 0:37