Linked Questions

4
votes
6answers
16k views

Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?

Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$? It is said that this is derived from trigonometry, but I couldn't find why this is the case.
7
votes
4answers
1k views

What is the value of $\arctan(1/2)+\arctan(1/5)+\arctan(1/8)$?

What is the value of : $$\arctan(1/2)+\arctan(1/5)+\arctan(1/8)?$$ I tried to do geometric solution:: Where in the angles we are looking for are shown, but I can't solve it. Can we use it with ...
4
votes
3answers
635 views

Partial Derivative of arctan

Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$ Find $f_x(x,y)$ My attempt, $$ \begin{aligned} f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\...
2
votes
2answers
3k views

Infinite sum of $\cot^{-1} (n^2 + 3/4)$.

I am trying to find the infinite sum $$\sum_{n=1}^\infty \cot^{-1} (n^2 + ( \frac{3}{4})),$$ I tried to get a telescopic series but I couldn't find one.
1
vote
3answers
3k views

Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.

Given $$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$ $$a,b \in (0,\frac{\pi}{2})$$ Show that $$a+2b=\frac{\pi}{4}$$ Does exist any faster method of proving that, ...
0
votes
8answers
181 views

Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$

First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine: $$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{...
2
votes
3answers
386 views

Find the values of $x$ such that $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$.

Find the values of $x$ such that $$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$. Checking for $x\in [-1,1]$ In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)...
1
vote
2answers
3k views

Derive the conditions $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ and $xy>-1$ for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$

$$ \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \text{, }xy<1\\ \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \text{, }xy>-1 $$ But, How do I reach the conditions $xy<1$ for the first ...
3
votes
2answers
830 views

Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$

Find $\tan x$ if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$ First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$ So $$\...
6
votes
4answers
226 views

What is the value of $\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)?$

I was playing about with some numbers when I came up with this fun question. What is the value of $\arctan \left(\frac xy\right) +\arctan \left(\frac yx\right)?$ Here is my method: As is ...
0
votes
2answers
185 views

How do I find a Taylor series of $\arctan \frac{2-2x}{1+4x}$ at $x=0$

I've been trying to find a Taylor series of $\arctan \frac{2-2x}{1+4x}$ at $x=0$ The only thing I could think of was trying to find formula for the $n$th derivative but was unable to find it so it ...
-1
votes
4answers
139 views

If $\tan A=-\frac 1 2$ and $\tan B =-\frac 1 3$, then $A+B =?$

Answer should be in radians Like π/4 (45°) π(90°). I used $\tan(A+B)$ formula and got $5/7$ as the answer, but that's obviously wrong.
5
votes
2answers
92 views

Find all $x$ such that $\sin x = \frac{4}{5}$ and $\cos x = \frac{3}{5}$.

Let $$ \left\{ \begin{array}{c} \sin x = \frac{4}{5} \\ \cos x = \frac{3}{5} \end{array} \right. $$ Find all of the possible values for $x$. My try: By dividing the equations we obtain $\...
1
vote
3answers
303 views

If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$

If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$ $\bf{My\; Try::}$ Given $$ \alpha ...
0
votes
4answers
1k views

Solve for $x$: $\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$

$$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$ $$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$ $$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$ $$6x^2 + 5x -1 = 0$$ $$(6x-1)(x+1)=0$$ ...

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