Is this a correct way to prove that  $$ \arctan(\mathrm{e}^{-v}) = \text{arccot}(\mathrm{e}^v) $$  $$\tan x = \frac{1}{\mathrm{e}^v} $$  Turn $\tan x$ into $\sin x$ and $\cos x$ $$\frac{\sin x}{\cos x} =  \frac{1}{\mathrm{e}^v}$$ Multiply $\cos x$ $$ \sin x = \frac{\cos x}{\mathrm{e}^v} $$ $$ \mathrm{e}^v \sin x = \cos x$$ $$ \mathrm{e}^v = \frac{\cos x}{\sin x}$$ Multiply by $\mathrm{e}^v$ and divide by $\sin x$ $$ \mathrm{e}^v = \cot x$$ $$\text{arccot}(\mathrm{e}^v) = x$$
17448
In my textbook it asks for me to:  Prove that there is no constant $C$ such that $\text{arccot}(x) - \text{arctan}(\frac{1}{x}) = C $ for all $x \ne 0$. Explain why this does not violate the zero-derivative theorem.  But I believe I have found such a $C$, i.e. $C =0$! I even asked WolframAlpha () which corroborates my answer.    This question appears in Apostol's Calculus Volume I, Second Edition: Exersize 6.22-11b  Edit: Mathematica's definition of arccot is different from the one in my textbook. Apostol's arccot maps a real number into $(0, \pi)$ while Mathematica's maps a real number into $(-\pi/2, \pi/2)$ Here they are super-imposed:
11215
I have two questions:  A. I know the taylor series for $\arctan(x)$ and for $e^x$. How do I find the series for $\arctan(x)\cdot e^x$ ?  B. Say I want to find the series for $\arctan(g(x))$, do I just subtitue $x$ with $g(x)$, Or do I have to start the process of finding the series from the beginning?  Thanks!
52787
How do I show this?  $$\cos^2(\arctan(x))=\frac{1}{1+x^2}$$  I have absolutely no idea. Thank you.
87145
Do you see any nice way to prove that $$\arctan\frac{\pi}{2}\ge1 ?$$  Thanks!  Sis.
55345
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.   Attempt:  I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.
45200
How does that bit work?  How is  $$\cos(\arcsin(x)) = \sin(\arccos(x)) = \sqrt{1 - x^2}$$
20127
My try, using $x = \sec(u)$ substitution:  $$ \begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &amp;=&amp; \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &amp;=&amp; \int \tan^2(u) \mathrm{d}u \\ &amp;=&amp; \tan(u) - u + C \\ &amp;=&amp; \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray} $$  However, according to , the answer should be: $$ \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C $$ When I derive this last answer I don't get back the integrand, but rather: $$ \frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)} $$  I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.  Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
41148
I'm doing some math exercises but I got stuck on this problem.  In the book of Bogoslavov Vene it says to prove that:  $$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.$$  It is easy if we do it like this: $\sin^2 x = (1-\cos x)(1+\cos x)=1-\cos^2 x$. But how to prove it just by going from the left part or from the right  Can anybody help me?  Thank you!
1937
How to prove for $x&gt;0$, $\dfrac{x}{1+x^2}&lt;\arctan x&lt;x$?   I started with saying $0&lt;\arctan x&lt;\frac{\pi}{2}$  when $x&gt;0$  I'm just not sure how to proceed with this proof, it works when I take random  values but I cant see how to prove that it works for any $x$.
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?
48681
$$\frac {2\tan {x}}{1-(\tan {x})^2} = (\sin {2x})^2$$  I tried a lot but I get nowhere
31956
I was reading a trig book in which the author had deduced the following formulas from a drawing. I am mentioning those as follows...     By looking at the above drawing the author deduced these formulas for $(\theta+90^\circ)$...  1. $\sin (\theta + 90^\circ)=\cos\theta$  2. $\cos (\theta + 90^\circ)=-\sin\theta$  3. $\tan (\theta + 90^\circ)=-\cot\theta$  I understood how did he conclude above formulas from the drawing.  As you can see the drawing depicts author's approach to derive the $(\theta+90^\circ)$ formulas by taking a similar triangle in the second quadrant, in the same way, to derive $(\theta-90^\circ)$ formulas, he advised to take a similar triangle (as in the drawing) in quadrant 4 as well. But, to the contrary he mentioned a trick to get those formulas which I do not understand.   The author told to replace $\theta$ in formulas 1,2 and 3 by $(\theta-90^\circ)$. So what I got by doing that is...  1. $\sin ((\theta-90^\circ) + 90^\circ)=\cos(\theta-90^\circ)$  2. $\cos ((\theta-90^\circ) + 90^\circ)=-\sin(\theta-90^\circ)$  3. $\tan ((\theta-90^\circ) + 90^\circ)=-\cot(\theta-90^\circ)$  By evaluating them further I get...  1. $\sin (\theta)=\cos(\theta-90^\circ)$ (I got this one correct)  2. $\cos (\theta)=-\sin(\theta-90^\circ)$ (wrong)  3. $\tan (\theta)=-\cot(\theta-90^\circ)$ (wrong)  But the author says that the outcome of the formulas should be...  1.$\sin (\theta)=\cos(\theta-90^\circ)$ (correct.)  2. $-\cos (\theta)=\sin(\theta-90^\circ)$   3. $\tan (\theta-90^\circ) =-\cot(\theta)$   I don't know where am I going wrong. The author told to replace $\theta$ by $(\theta-90^\circ)$ then why am I getting my answers incorrect? Where is the flaw? Please help me to track it down.
63583
My (very simple) question to a friend was how do I prove the following using basic trig principles:  $\tan75^\circ = 2 + \sqrt{3}$  He gave this proof (via a text message!)  $1. \tan75^\circ$   $2. = \tan(60^\circ + (30/2)^\circ)$  $3. = (\tan60^\circ + \tan(30/2)^\circ) / (1 - \tan60^\circ \tan(30/2)^\circ) $  $4. \tan (30/2)^\circ = \dfrac{(1 - \cos30^\circ)}{ \sin30^\circ}$    Can this be explained more succinctly as I'm new to trigonometry and a little lost after (2.) ?    EDIT  Using the answers given I'm almost there:   $\tan75^\circ$    $\tan(45^\circ + 30^\circ)$   $\sin(45^\circ + 30^\circ) / \cos(45^\circ + 30^\circ)$   $(\sin30^\circ.\cos45^\circ + \sin45^\circ.\cos30^\circ) / (\cos30^\circ.\cos45^\circ - \sin45^\circ.\sin30^\circ)$    $\dfrac{(1/2\sqrt{2}) + (3/2\sqrt{2})}{(3/2\sqrt{2}) - (1/2\sqrt{2})}$   $\dfrac{(1 + \sqrt{3})}{(\sqrt{3}) - 1}$   multiply throughout by $(\sqrt{3}) + 1)$   Another alternative approach:   $\tan75^\circ$    $\tan(45^\circ + 30^\circ)$   $\dfrac{\tan45^\circ + \tan30^\circ}{1-\tan45^\circ.\tan30^\circ}$   $\dfrac{1 + 1/\sqrt{3}}{1-1/\sqrt{3}}$   at point 6 in above alternative
84821
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$   I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution too, but not pretty . !
90904
I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.      $$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$     My Attempt:  Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$.     I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this
38838
Some years ago, someone had shown me the formula (1). I have searched for its origin and for a proof. I wasn't able to get true origin of this formula but I was able to find out an elementary proof for it.  Since then, I'm interested in different approaches to find more formulae as (1).  What other formulas similar to ($1$) are known?  Two days ago, reading the book of Lewin "Polylogarithms and Associated Functions" I was able to find out formula (2).  $\displaystyle \dfrac{1}{3}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{2-x}\right)dx\tag1$   $\displaystyle \dfrac{2}{5}C=\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{\sqrt{5}x(1-x)}{1+\sqrt{5}-\sqrt{5}x}\right)dx-\int_0^1 \dfrac{1}{x}\arctan\left(\dfrac{x(1-x)}{3+\sqrt{5}-x}\right)dx\tag2$   $C$ being the Catalan's constant.  I have a proof for both of these formulae.  My approach relies on the following identity:  For all real $x&gt;1$,  $\displaystyle \int_0^1 \dfrac{1}{t} \arctan \left (\dfrac{t(1-t)}{\frac{x+1}{2}-t}\right) dt=\int_1^{\frac{\sqrt{x}+1}{\sqrt{x}-1}}\dfrac{\log(t)}{1+t^2}dt$
77505
Could anyone offer a proof that  $$ \cos^2(x)\cosh^2(y) + \sin^2(x)\sinh^2(y) = -1 + \sin^2(x) - \sinh^2(y)? $$