I know that the series representation of $\tan(x)$ is:  $x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}...$  I can derive this by using the fact that $\tan(x)=\dfrac{\sin(x)}{\cos(x)}$. But I don't know how to find the general sigma formula for this function.  I have seen the general formula here:   But I don't know how the Bernoulli number is involved. Can you explain this further so I can understand?
122835
I found the following formula here: .  Taylor series of $\tan x$:  $$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} $$.  I do not understand what the Bernoulli number means in this formula. My problems are the following ones:  I do not know anything about Bernoulli numbers. I do not know how to expand the Bernoulli number in the formula you see further up. What expansion will be the final result? How did you arrive there? Why did you arrive there? How did you know that you should use the specific Bernoulli-related terms that you used in your answer? What is the purpose of the Bernoulli number in this formula?  A great deal of thanks.
77357
This problem has baffled me. I have tried using $t = \tan(x/2)$ identities but end up with a 4 th order polynomial I cant solve.
101244
Testing with my calculator in degree mode, I have found the following to be true:  $$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$  Why is this? What is the proof or explanation?
93029
How do we prove that $$I(m)=\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x=\frac{\log{m}}{m^2-1}$$  I see that $$I(m)=\frac{\partial}{\partial m} \int_{0}^{\pi/2} \arctan({m\tan x}) \ \mathrm{d}x$$  But I don't see how to use this fact. Can we? Please help me out, and if possible please post a solution using differentiation under the integral sign.
84672
Letting $x=\tan(u)$, $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{1+\tan(u)^2}\sec^2(u) \, du=u+k=\arctan(x)+k$$ Also,  $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{(x-i)(x+i)} \, dx=\frac{1}{2i}\int\frac{1}{(x-i)}-\frac{1}{(x+i)} \, dx$$ $$=\frac{1}{2i}\left(\ln(x-i)-\ln(x+i)\right)+c$$ $$=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+c$$ Giving $$\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+q$$  Why is this correct? What is the nature of $q$ (is it 'flexible' so the equality doesn't really mean much)?  I think it probably has something to do with the relationship between $\log(z)$ and $\arg(z)$, but $\arg(z\pm i)$ is hard to calculate neatly.
64103
Our teacher assigned extra credit for which we are allowed to use any source (including this website) to find the answer. This material has not been covered in class or in the book we are using. We learned Taylor series using series of real numbers but we have not covered anything remotely similar to the problem below so I have no idea how to even begin this problem.   Let i =$\sqrt{−1}$   a) Use the Taylor series centered at x=0 to prove that, for all $\theta$,  $e^{i\theta} = cos(\theta) + isin(\theta)$  b) Use above answer to prove that $[cos(\theta)+ isin(\theta)]^n = cos(n\theta)+ isin(n\theta)]$  c) We may write $[cos(\theta)+ isin(\theta)]^n = cos^n(x)(1+itan(\theta))^n$  Use the 5th order Taylor polynomial for $(1+x)^\frac{1}{3}$ with $x = itan(\theta)$ to find an estimate for $sin\left(\frac{\theta}{3}\right)$.  Use this approximation to estimate the value of $sin\left(\frac{\pi}{18}\right)$.  Hint: Two complex numbers $a+bi$ and $c+di$ are equal if $a=c$ and $b=d$.
24894
Show that,   $$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$  There is a mixed of sin and tan, how  can I simplify this to $\frac{\pi}{4}$  We know the identity of $\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)$
39231
$$ \arctan\left(\frac{1}{1+n+n^2}\right)$$  My professor wrote this as  $$\arctan(n+1) - \arctan(n)$$   I don't understand how this expression is right?
73802
I'm working on a problem:  Compute cos(5°) to 5 decimal places with Maclaurin's Series  I know that that function cos(x) has a Mclaurin representation of:  $ \sum_{n=0}^\infty \frac{(-1)^n (x)^{2n}}{(2n)!} $  The only part I am unsure of is how to find out many terms n to choose.  I'm not sure if I am supposed to use the alternating series estimation or taylor series estimation or what.
115154
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$   So, LS= $$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$ $$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$ $$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$ Now, considering the fact that I must have a common denominator to subtract, would this be correct: $$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$ I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
116770
I know that if $x$ is a rational multiple of $\pi$, then $\tan(x)$ is .  Is there a fairly simple way to express $x$ as $\pi\frac{m}{n}$, if $\tan(x)$ is given as a square root of a rational?
102137
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$   My approach :   Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then  $$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$  But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
24692
Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$: $$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$  The question isn't homework or anything, just a thought tease. I tried for a long while but couldn't find anything remotely close. Thanks in advance for the help.
23502
How does one get from this $$\zeta(s)=\sum_{k=1}^{\infty}\frac1{k^s}$$ to the integral representation $$\zeta(s)=\frac1{\Gamma(s)}\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$ of the Riemann Zeta function?  I can see that it can be rewritten as  $$\Gamma(s)\zeta(s)=\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$ and the Gamma function as an integral yields  $$\zeta(s)\int_{0}^\infty \frac{x^{s-1}}{e^x}dx=\int_{0}^\infty \frac{x^{s-1}}{e^x-1}dx$$  But this approach does not work as the right integral does not converge.  So how does one go from the summation to the integral representation?
114165
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$   $\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$  Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get  $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$  So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$  So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$  My question is how can we solve it in some shorter way, Help required, Thanks
72748
Trig functions are simple ratios, but what do $\cosh$, $\sinh$ and $\tanh$ compute? How are they related to Euler’s number anyway?
54168
I tried in this way.$\tan(a+b)=\frac{(\tan a + \tan b)}{1 - \tan a \tan b }$value of $\tan \frac{\pi}{7}$ is coming in decimal.what to do