How to simplify $\mathrm{\Re}[\sqrt{2}{\mathrm{tan}}^{-1}\frac{x}{\sqrt{i}}]$?

Yesenia Sherman
2022-06-24
Answered

How to simplify $\mathrm{\Re}[\sqrt{2}{\mathrm{tan}}^{-1}\frac{x}{\sqrt{i}}]$?

You can still ask an expert for help

enfujahl

Answered 2022-06-25
Author has **20** answers

$\mathrm{\Re}[\sqrt{2}{\mathrm{tan}}^{-1}\frac{x}{\sqrt{i}}]=\mathrm{\Re}[\sqrt{2}{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4}]=\frac{(\sqrt{2}{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4})+(\sqrt{2}{\mathrm{tan}}^{-1}x\cdot {e}^{i\pi /4})}{2}$

$=\frac{1}{\sqrt{2}}({\mathrm{tan}}^{-1}x\cdot {e}^{i\pi /4}+{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4})=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{x\cdot {e}^{i\pi /4}+x\cdot {e}^{-i\pi /4}}{1-x\cdot {e}^{i\pi /4}\cdot x\cdot {e}^{-i\pi /4}}$

$=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{2x\mathrm{cos}\frac{\pi}{4}}{1-{x}^{2}}=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{\sqrt{2}x}{1-{x}^{2}}$

where line $2\to 3$ is achieved by using that $\mathrm{tan}(x+y)=\frac{\mathrm{tan}x+\mathrm{tan}y}{1-\mathrm{tan}x\mathrm{tan}y}$

The differs from the expression you give by a constant, after further noting that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$ for x>0

$=\frac{1}{\sqrt{2}}({\mathrm{tan}}^{-1}x\cdot {e}^{i\pi /4}+{\mathrm{tan}}^{-1}x\cdot {e}^{-i\pi /4})=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{x\cdot {e}^{i\pi /4}+x\cdot {e}^{-i\pi /4}}{1-x\cdot {e}^{i\pi /4}\cdot x\cdot {e}^{-i\pi /4}}$

$=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{2x\mathrm{cos}\frac{\pi}{4}}{1-{x}^{2}}=\frac{1}{\sqrt{2}}{\mathrm{tan}}^{-1}\frac{\sqrt{2}x}{1-{x}^{2}}$

where line $2\to 3$ is achieved by using that $\mathrm{tan}(x+y)=\frac{\mathrm{tan}x+\mathrm{tan}y}{1-\mathrm{tan}x\mathrm{tan}y}$

The differs from the expression you give by a constant, after further noting that ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$ for x>0

asked 2021-08-20

Let P(x, y) be the terminal point on the unit circle determined by t. Then

asked 2022-04-21

If $0<{\alpha}_{1}<{\alpha}_{2}<\cdots <{\alpha}_{n}<\frac{\pi}{2}$ is given then prove: : $\mathrm{tan}\alpha}_{1}<\frac{{\mathrm{sin}\alpha}_{1}+\cdots +{\mathrm{sin}\alpha}_{n}}{{\mathrm{cos}\alpha}_{1}+\cdots +{\mathrm{cos}\alpha}_{n}}<{\mathrm{tan}\alpha}_{n$

asked 2022-03-23

Why can't $\frac{1}{\mathrm{sin}x\mathrm{cos}x}$ be expressed in the form $\frac{A}{\mathrm{sin}x}+\frac{B}{\mathrm{cos}x}$

$\frac{1}{\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)}=\frac{A}{\mathrm{sin}x}+\frac{B}{\mathrm{cos}x}\Rightarrow 1=A\mathrm{cos}x+B\mathrm{sin}x$

I let$x=\frac{\pi}{2}$ , getting A=-1. Then I let $x=\pi$ , getting B=1. This means that $1=\mathrm{cos}x-\mathrm{sin}x$ which is obviously wrong most of the time.

I let

asked 2022-03-28

Computing ${\int}_{0}^{2\pi}{(1+2\mathrm{cos}t)}^{n}\mathrm{cos}ntdt$

asked 2022-03-30

Prove the following trig identities:

$2{\mathrm{sin}}^{2}\theta -1={\mathrm{sin}}^{2}\theta -{\mathrm{cos}}^{2}\theta$

${\mathrm{cos}}^{2}t={\mathrm{sin}}^{2}t+2{\mathrm{cos}}^{2}t-1$

asked 2022-03-30

Prove that $\sum _{n\in \mathbb{N}}{\mathrm{sin}}^{3}\left\{\frac{2}{n}\right\}$ converges

asked 2022-07-27

A water tower is located 325 ft. from a building. From a window in the building it is observed that the angle of elevation to the top of the tower is 39 degrees and the angle of depressionto the bottom of the tower is 25 degrees. How tall is thetower? How high is the window?