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

I am working on a problem in Partha Mitra's book Observed Brain Dynamics (the problem was originally from Rudin's textbook Real and Complex Analysis, and appears on page 54 of Mitra's book). Unfortunately, the book I have does not contain any solutions... Here is the question:

By considering the first few terms of the series expansion of sin(x) and cos(x), show that there is a real number $x_0$ between 0 and 2 for which $\cos(x_0)=0$ and $\sin(x_0)=1$. Then, define $\pi=2x_0$, and show that $e^{i\pi/2}=i$ (and therefore that $e^{i\pi}=-1$ and $e^{2\pi i}=1$.

Attempt at a solution: In a previous problem I derived the series expansions for sine and cosine as

$$ \sin(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{x^{2n+1}}{(2n+1)!} \right) \right] $$

$$ \cos(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{x^{2n}}{(2n)!} \right) \right] $$

My thought is that you can show that $\cos(0)=1$ (trivially), and that $\cos(2)<0$ (less trivially). This then implies that there is a point $x_0$ between 0 and 2 where $\cos(x_0)=0$, since the cosine function is continuous. However, I do not understand how you could then show that $\sin(x_0)=1$ at this same point. My approach may be completely off here.

I believe that the second part of this problem ("Then, define $\pi=2x_0$...") will be easier once I get past this first part.

Thanks so much for the help. Also - I swear this is not a homework assignment. I am reading through this book on my own to improve my math.

share|cite|improve this question
I think the hint is to consider $\sin$ and $\cos$ not in terms of their series expansions, but as solutions to the differential equation $y'' + y = 0$. There's a pretty detailed discussion of this in George Simmons' book Differential Equations with Historical Notes. I will try to find the page number if you are interested. – MJD Sep 12 '12 at 2:39
The Simmons discussion, which you may enjoy, is on pages 115–118 of the book I mentioned. Among other familiar properties, he derives the property $\sin^2 x + \cos^2 x = 1$ from the differential equation I mentioned above. Since the differential equation property is implied by the power series for $\sin$ and $\cos$, you might find this interesting. – MJD Sep 12 '12 at 2:46
Thanks for the input! I'll try to check that out! – Alex Sep 12 '12 at 2:56
The development of $\pi$ that you are discussing appears on pages 182–183 of Principles of Mathematical Analysis by W. Rudin. It is part of the main text, not an exercise. Although I wouldn't say it is explained in detail, the development is sufficiently clear that it can be followed. If you're still interested but puzzled, you should have a look at it. – MJD Mar 19 '13 at 4:38
up vote 0 down vote accepted

How to show that $\sin (x_0)=1$ if $\cos (x_0)=0$? Quite simply:

$$\sin^2 x+\cos^2 x=1$$

(you may also want to specify that $\sin x$ is positive in the given range)

share|cite|improve this answer
But the question said By considering the first few terms of the series expansion of sin(x) and cos(x), show that [..] – user2468 Sep 12 '12 at 2:44
Yeah thats a good point... – Alex Sep 12 '12 at 2:46
I think you must be right Julien - you can show that cos(x)=0 by using the first few terms, and then use that identity you mentioned (which I derived in a separate problem). Thanks for the help! – Alex Sep 12 '12 at 2:51
@JenniferDylan Hm, yes I read it rather quickly. I understood the and as an and hence ... but now I see why this could be labeled as misinterpretation. It just seems doing them separately is complicating things for no good reason. – user39572 Sep 12 '12 at 2:56

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.