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$$ 4\operatorname{arccot}(2)+\arctan\left(\frac{24}7\right)=\pi $$ original image

To prove the above result, I tried to equate the original expression to some constant $a$ such that $0<a<2.5\pi$ (from the range of the inverse tangent). When I try to solve for $a$ by taking the tangent or sine of both sides, I arrive at the equations:

$$\begin{align} \sin(a) &=0 \\ \tan(a) &=0 \end{align}$$ which gives me two solutions ($\pi$ and $2\pi$) within the specified range.

I have already seen other solutions using complex numbers, so I would really appreciate if someone could point out where I'm going wrong rather than a solution via another method.

Blue
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Vulgar Mechanick
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    You should have $\sin a = 0, \tan a = 0$. Could you please show us how you got to that result? Also, the range of $\arctan x$ is actually $[-\frac{\pi}{2}, \frac{\pi}{2}]$. – Toby Mak May 30 '20 at 12:33
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    Hint: $\operatorname{arccot}(2)=\arctan\left(\frac12\right)$ and the formula for the sum of tangents gives $\tan(2\arctan(x))=\frac{2x}{1-x^2}$ – robjohn May 30 '20 at 12:46
  • Toby Mak, Im extremely sorry for the typo, that is exactly what I got. I got the upper bound from the fact that the expression is of the form 4arctanA + arctanB where A and B are both positive. So a has to be less than 5π/2. – Vulgar Mechanick May 30 '20 at 12:52
  • Robjohn. I Know but that gives me 2 π as in incorrect possible solution when i take the inverse tangent of 0 at the end – Vulgar Mechanick May 30 '20 at 12:55
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    @OVERWOOTCH: pay attention to where your angle is $\arctan\left(\frac12\right)\in\left(0,\frac\pi4\right)$, so $\tan\left(2\arctan\left(\frac12\right)\right)=\frac43$ and $2\arctan\left(\frac12\right)\in\left(\frac\pi4,\frac\pi2\right)$ so $2\arctan\left(\frac12\right)=\arctan\left(\frac43\right)$. This means that $4\arctan\left(\frac12\right)\in\left(\frac\pi2,\pi\right)$ and $\tan\left(4\arctan\left(\frac12\right)\right)=-\frac{24}7$. This means that $4\arctan\left(\frac12\right)=\pi-\arctan\left(\frac{24}7\right)$. – robjohn May 30 '20 at 13:00
  • Where does this question come from? – Toby Mak Jun 14 '21 at 04:27

2 Answers2

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You need to pay attention to where your angle is. Note that $$ \operatorname{arccot}(2)=\arctan\left(\frac12\right)\tag1 $$ and that $\arctan\left(\frac12\right)\in\left(0,\frac\pi4\right)$. The identity $\tan(2\arctan(x))=\frac{2x}{1-x^2}$ says $$ \tan\left(2\arctan\left(\frac12\right)\right)=\frac43\tag2 $$ and $2\arctan\left(\frac12\right)\in\left(\frac\pi4,\frac\pi2\right)$ so $$ 2\arctan\left(\frac12\right)=\arctan\left(\frac43\right)\tag3 $$ Thus, $4\arctan\left(\frac12\right)\in\left(\frac\pi2,\pi\right)$ and $$ \begin{align} \tan\left(4\arctan\left(\frac12\right)\right) &=\tan\left(2\arctan\left(\frac43\right)\right)\tag4\\ &=-\frac{24}7\tag5 \end{align} $$ Therefore, $$ 4\arctan\left(\frac12\right)=\pi-\arctan\left(\frac{24}7\right)\tag6 $$ Putting together $(1)$ and $(6)$ gives $$ 4\operatorname{arccot}(2)+\arctan\left(\frac{24}7\right)=\pi\tag7 $$

robjohn
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    Im sorry; Im new to this forum and a little unfamiliar with the mathematical notation. How did you arrive at 2arctan(1/2) "∈ (pi/4, pi/2) ( Im assuming it means "lies in the interval"). Also could you elaborate on why I cant just take the tangent of both sides and take the inverse tangent at the end? – Vulgar Mechanick May 30 '20 at 13:23
  • As I said, you need to pay attention to where your angle is. Although we know that $\tan\left(2\arctan\left(\frac12\right)\right)=\frac43$, we cannot immediately say that $2\arctan\left(\frac12\right)=\arctan\left(\frac43\right)$, without knowing $2\arctan\left(\frac12\right)\in\left(-\frac\pi2,\frac\pi2\right)$. Since $\arctan\left(\frac12\right)\le\arctan(1)=\frac\pi4$, we know that $2\arctan\left(\frac12\right)\le\frac\pi2$. Thus, we can now say that $2\arctan\left(\frac12\right)=\arctan\left(\frac43\right)$. – robjohn May 30 '20 at 14:06
  • Furthermore, we know that $2\arctan\left(\frac12\right)=\arctan\left(\frac43\right)\ge\arctan(1)=\frac\pi4$. Therefore, we know that $2\arctan\left(\frac12\right)\in\left(\frac\pi4,\frac\pi2\right)$. – robjohn May 30 '20 at 14:06
  • But in the same way, we also know that arctan(1/2)<arctan(1/sqrt3). That means we can further lower the upper bound of arctan(1/2) to pi/6 if I understood you correctly? – Vulgar Mechanick May 30 '20 at 14:23
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    Yes, that is correct. However, for this problem, we are only interested in determining the quadrant of $4\arctan\left(\frac12\right)$. – robjohn May 30 '20 at 16:32
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Hint:

Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

and find $$(2+i)^4(7+24i)=\cdots-7^2-24^2$$

Find arguments in both sides

lab bhattacharjee
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  • (+1) I like using complex multiplication to show that $\overbrace{(1+i\tan(A))}^{\arg=A}\overbrace{(1+i\tan(B))}^{\arg=B}\overbrace{(1+i\tan(C))}^{\arg=C}\in\mathbb{R}$ if $A+B+C=\pi$ to show that $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$ for the angles in a triangle. – robjohn May 30 '20 at 17:06
  • However, it should be pointed out that $4\arctan(x)+\arctan(y)\lt\frac{5\pi}2$ to ensure that the argument is not actually $3\pi$. – robjohn May 30 '20 at 17:18