Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can we find the exact value(not numerical/approximation) of sin 1? I tried to do so by solving a cubic equation using Cardano formula but I ended up with complex nested radicals): I was told to use other methods but that won't yield exact values

share|cite|improve this question
Wouldn't its exact value be $\sin(1)$? – AnonSubmitter85 Aug 25 '13 at 7:24
If the angle is in degree, have a look into – lab bhattacharjee Aug 25 '13 at 7:25
Isnt sin(1) an irrational? – al-Hwarizmi Aug 25 '13 at 7:26
I mean without computer software, how do you find it manually?Can you show me? – Tom Lynd Aug 25 '13 at 7:27
@ al Hwarizmi yes it is but can't it be represented in a form like the multiples of 3 can be... – Tom Lynd Aug 25 '13 at 7:29

sin(1) is a Transcendental number.

ie ,sin(1) is not a solution a non-zero polynomial equation with rational coefficients.In other words sin(1) can't be expressed in closed form only using fractions and radicals. Note: sin(a), where a is algebraic(not transcendental) ,is transcendental.

Does that help you?

share|cite|improve this answer
@TomLynd , I didn't really understand what you meant by their solutions being incorrect. The links are for info on transcendental number and for the reason why sin(1) is transcendental. What I meant by sin(1) being not the solution of...... is that sin(1) can't be a solution of a polynomial equation with non-zero rational coefficients. There is no closed expression which has only radicals,fractions and algebraic numbers. There might be some expressions which involve transcendental numbers like e and $\pi$. – jaseem Aug 25 '13 at 12:35
$\sin (1) = \Im(e^i)$ – Mats Granvik Aug 25 '13 at 12:51
@ jaseem you better have a look at – lab bhattacharjee – Tom Lynd Aug 25 '13 at 13:55
@TomLynd that is for one degree , or for $sin(\frac{\pi}{180})$ and $\frac{\pi}{180}$ is not algebraic.But 1(1 radian) is algebraic and therefore sin(1) is transcendental. – jaseem Aug 25 '13 at 15:04
Series[Sin[x], {x, 0, 14}]

$$\sin (x) = x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{39916800}+\frac{x^{13}}{6227020800}+O\left(x^{15}\right)$$

DeleteCases[CoefficientList[Series[Sin[x], {x, 0, 14}], x], 0]

$$\sin (1) = 1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}+\frac{1}{362880}-\frac{1}{39916800}+\frac{1}{6227020800}-...$$

$$1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}+\frac{1}{362880}-\frac{1}{39916800}+\frac{1}{6227020800}-... = 0.84147098480866...$$

$$\sin (1) = 0.84147098480790...$$

Table[(2*n - 1)!, {n, 1, 7}]

{1, 6, 120, 5040, 362880, 39916800, 6227020800}

share|cite|improve this answer
He wants a closed form. My comment addresses this. – jaseem Aug 25 '13 at 12:26

If $1$ is degrees here's 2 closed forms:

$$\frac{1}{2}ie^{-\frac{i\pi}{180}} - \frac{1}{2}ie^{\frac{i\pi}{180}}$$

$$-\frac{1}{2}(\sqrt[180]{-1} - 1)(\sqrt[180]{-1} + 1)(-1)^{\frac{89}{180}}$$

If $1$ is radians here's one closed form:

$$\frac{i e^{-i}}{2}-\frac{i e^i}{2}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.