English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-pi/2 <arctan(y)<0

Solution

- \frac{π}{2} < arctan (y) < 0

y < 0

Interval Notation

(- \infty, 0)

Solution steps

- \frac{π}{2} < arctan (y) < 0

a < u < b

then

a < u and u < b

- \frac{π}{2} < arctan (y) and arctan (y) < 0

- \frac{π}{2} < arctan (y) :

True for all

y \in R

- \frac{π}{2} < arctan (y)

Switch sides

arctan (y) > - \frac{π}{2}

Range of

arctan (y) : - \frac{π}{2} < arctan (y) < \frac{π}{2}

Function range definition

The range of the basic

arctan

function is

- \frac{π}{2} < arctan (y) < \frac{π}{2}

- \frac{π}{2} < arctan (y) < \frac{π}{2}

arctan (y) > - \frac{π}{2} and - \frac{π}{2} < arctan (y) < \frac{π}{2} : - \frac{π}{2} < arctan (y) < \frac{π}{2}

Let

y = arctan (y)

Combine the intervals

y > - \frac{π}{2} and - \frac{π}{2} < y < \frac{π}{2}

Merge Overlapping Intervals

y > - \frac{π}{2} and - \frac{π}{2} < y < \frac{π}{2}

The intersection of two intervals is the set of numbers which are in both intervals

y > - \frac{π}{2}

and

- \frac{π}{2} < y < \frac{π}{2}

- \frac{π}{2} < y < \frac{π}{2}

- \frac{π}{2} < y < \frac{π}{2}

True for all y

True for all y \in R

arctan (y) < 0 : y < 0

arctan (y) < 0

arctan (x) < a

then

x < tan (a)

y < tan (0)

tan (0) = 0

tan (0)

Use the following trivial identity:

tan (0) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

y < 0

Combine the intervals

True for all y \in R and y < 0

Merge Overlapping Intervals

True for all y \in R and y < 0

The intersection of two intervals is the set of numbers which are in both intervals

True for all

y \in R

and

y < 0

y < 0

y < 0

sin (θ) = \frac{3}{2} and tan (θ) < 0

sin (x) 0 < x < π

0 \leq sin (x) \leq 1

- \frac{π}{2} < arcsin (y) < \frac{π}{2}

sin (θ) = - \frac{1}{2} and cos (θ) > 0