The trigonometric functions for half angles have well-defined and simple rules. For example $$\tan\left(\frac x2\right)=\frac{\sin x}{1+\cos x}=\frac1{\csc x+\cot x}$$ But I can't get the exact value for some expressions as simple as $\tan(\pi/8)$. No matter what I tried, Mathematica wouldn't understand that it's equal to $\sqrt2-1$, for example:

Simplify[Tan[Pi/8] + 1 - Sqrt[2]]

does nothing. I tried TrigReduce, TrigExpand, TrigFactor, FullSimplify and some other unrelated functions and they all failed to give the exact value of $\tan(\pi/8)$. This became ironic when I checked some other angles such as $\pi/12$ and turned out it does know e.g. $\sin(\pi/12)$ whose only way of calculation is apparently, through half angle formulas.

Why this happens and how can I get Mathematica to calculate such values when they do have an exact expression?

  • 1
    $\begingroup$ Of course, the question is a bit flawed because Tan[Pi/8] is an exact numerical constant. $\endgroup$
    – John Doty
    Oct 29, 2021 at 15:36
  • $\begingroup$ @JohnDoty please feel free to edit it if you can think of better words. My vocabulary is not that great. $\endgroup$
    – polfosol
    Oct 29, 2021 at 15:38
  • 1
    $\begingroup$ Well, what is it that you want, really? What problem are you trying to solve? $\endgroup$
    – John Doty
    Oct 29, 2021 at 15:40
  • 1
    $\begingroup$ Tan[Pi/8] // N // RootApproximant $\endgroup$
    – Bob Hanlon
    Oct 29, 2021 at 16:25
  • $\begingroup$ With version 12.3.1 for Microsoft Windows (64-bit) (June 19, 2021), FullSimplify[Tan[Pi/8] + 1 - Sqrt[2]] yields 0. $\endgroup$
    – bbgodfrey
    Oct 30, 2021 at 13:26

3 Answers 3


An answer to How does Mathematica calculate $\sin(\pi/5)$? by @J.M. points out that in the Notes on Internal Implementation, it is implied that FunctionExpand is the way to expand trigonometric functions in terms of radicals when possible:

FunctionExpand uses an extension of Gauss's algorithm to expand trigonometric functions with arguments that are rational multiples of π.

(*  Sqrt[(2 - Sqrt[2])/(2 + Sqrt[2])]  *)

The internal function that does what the OP wants is called Simplify`TrigToRealRadicals[], which is called by FunctionExpand and ToRadicals in this case. It in turn calls a function like System`TrigToRadicalsDump`tan[], which computes the conversion. These function memoize their results, which makes timing them a challenge. However, the timings below show that FunctionExpand does a lot of unnecessary work. We give a complicated example. At first, FunctionExpand does nothing, and ToRadicals returns the simplified expansion in terms of the complex exponential function.

(* Cos[π/257]  <-- takes <0.02s *)

(*  -(1/2) (-1)^(256/257) (1 + (-1)^(2/257))  <-- takes 0.004s *)

If the trig-to-radicals functionality has been loaded, then we get a different behavior. I cannot say for certain what functions cause the loading of the package. The first time I tested, it had been loaded and I got results like those below. The second, I retested with a fresh kernel and got the above, which confused me for a while.

Simplify`TrigToRealRadicals; (* load/initialize *)
System`TrigToRadicalsDump`cos[π/257]; // AbsoluteTiming
(*  {146.975, Null}  <-- the expansion is now memoized *)

ToRadicals[Cos[π/257]] // LeafCount
(*  263365  <-- in terms of (real) radicals; takes only 0.06s *)

Here's a speed comparison, assuming the initialization of System`TrigToRadicalsDump`cos[π/257] has been done.

Simplify`TrigToRealRadicals[Cos[π/257]]; // RepeatedTiming
ToRadicals[Cos[π/257]]; // RepeatedTiming
RootReduce[Cos[π/257]]; // RepeatedTiming
FunctionExpand[Cos[π/257]]; // AbsoluteTiming
  { 0.0579771, Null}
  { 0.0593911, Null}
  { 1.28168,   Null}
  {37.3825,    Null}
  • $\begingroup$ The formatting of third paragraph makes it hard to read/understand. By the way, is there anything special about number 257 or is it chosen randomly? $\endgroup$
    – polfosol
    Oct 30, 2021 at 19:28
  • $\begingroup$ @polfosol n=257 is a prime such that a regular n-gon can be constructed by ruler and compass. That comes down to writing Cos[Pi/257] in terms of square roots, more or less. The example is borrowed from the linked question. Such primes have to be Fermat primes, which have the form 2^(2^k)+1. (And thanks for alerting me to a missing backtick.) $\endgroup$
    – Michael E2
    Oct 30, 2021 at 20:26
TrigToExp[Tan[Pi/8]] // FullSimplify
-1 + Sqrt[2]

Or, even better:

Tan[Pi/8] // RootReduce
-1 + Sqrt[2]
  • $\begingroup$ Who would have thought, of all the functions, RootReduce would do the job? a strange world we live in... $\endgroup$
    – polfosol
    Oct 29, 2021 at 15:36
  • 6
    $\begingroup$ @polfosol You apparently wanted an explicit algebraic number, so RootReduce made perfect sense to me. $\endgroup$
    – John Doty
    Oct 29, 2021 at 15:38

Try to use ToRadicals[Tan[Pi/8]]//FullSimplify


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