So, I was playing around on Wolfram|Alpha (as we nerds like to do) and it said cos(1) was transcendental. Could someone provide me with the proof that cos(1) is transcendental?

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    $\begingroup$ I believe it comes from the fact that $\cos(1) = (e^i+e^{-i})/2$, so you're looking for the proof that $e^i$ is transcendental. $\endgroup$ Feb 16, 2014 at 0:19
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    $\begingroup$ See the Lindemann-Weierstrass theorem, in conjunction with Euler's formula. $\endgroup$
    – Lucian
    Feb 16, 2014 at 0:22

2 Answers 2


If $\cos1$ is algebraic, so is $\sin1=\sqrt{1-\cos^21}$. Thus $e^i=\cos1+i\sin1$ is algebraic. But, by Lindemann's theorem, $e^\alpha$ is transcendental whenever $\alpha$ is a nonzero algebraic number.


Since $ i $ is a non zero algebraic number then $ \lbrace i , 2i ,0\rbrace $ is a set of distinct algebraic numbers in $\mathbb{C}$. By the Lindemann-Weierstrass theorem for any non-zero algebraic numbers $ \beta_{1},\beta_{2},\beta_{3} $ we have $ \beta_1 e^{i} + \beta_2 e^{2i}+\beta_3 e^{0}\ne 0 $. To obtain a contradiction assume that $ cos(1) $ is algebraic. Recall that $ cos(1) =\dfrac{e^{i}+e^{-i}}{2}. $ Then $ 2cos(1) .e^{i}-e^{2i}-1=0 $ and hence $ (2cos(1) )e^{i}-1.e^{2i}-1.e^{0}=0 $. But this is a contradiction with the Lindemann-Weierstrass theorem. Therefore $ cos(1) $ is transcendental. $\square $


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