What is $\theta$ equal to

1. in radians,

2. in degrees,

if,$\mathrm{cot}\left(\theta \right)=1$

1. in radians,

2. in degrees,

if,

a2linetagadaW
2021-01-08
Answered

What is $\theta$ equal to

1. in radians,

2. in degrees,

if,$\mathrm{cot}\left(\theta \right)=1$

1. in radians,

2. in degrees,

if,

You can still ask an expert for help

oppturf

Answered 2021-01-09
Author has **94** answers

1. $\theta =\frac{\pi}{4}+\pi n$

2.$\theta ={45}^{\circ}+{180}^{\circ}n$

2.

asked 2021-12-27

Why is $\mathrm{tan}\theta \approx \frac{1}{\frac{\pi}{2}-\theta}$ for $\theta$ close to $\frac{\pi}{2}$ ?

I wanted to see what the behaviour of the steep part of the$\mathrm{tan}$ curve was like, i.e. the behaviour of $\mathrm{tan}\left(x\right)\text{}as\text{}x\to {\left(\frac{\pi}{2}\right)}^{-}$ . So by thinking about a shift of the graph of $\mathrm{tan}\left(x\right)\text{}by\frac{\pi}{2}$ to the left, I put some small (positive and negative) values of $\theta$ into my calculator for the function $\mathrm{tan}(\theta +\frac{\pi}{2})$

$\mathrm{tan}\theta \approx \frac{1}{\frac{\pi}{2}-\theta}$ for $\theta$ close to $\frac{\pi}{2}$ ?

or, in more colloquial terms,

The steep part of$\mathrm{tan}x$ is just like the steep part of $\frac{1}{x}$

But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of$\mathrm{tan}\left(x\right)$ . Is there a more intuitive explanation? I couldn't think of any explanations analogous to those explaining small angle approximations.

I wanted to see what the behaviour of the steep part of the

or, in more colloquial terms,

The steep part of

But why is this the case? I couldn't deduce it easily using the Maclaurin expansion of

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[Graph]
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For any $x\in [0,1]$ show that

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For any $x\in [0,1]$ show that

$\mathrm{arcsin}(x)+\mathrm{arccos}(x)=\frac{\pi}{2}$

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