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0<= arctan(x)<= pi/4

Solution

0 \leq arctan (x) \leq \frac{π}{4}

0 \leq x \leq 1

Interval Notation

[0, 1]

Solution steps

0 \leq arctan (x) \leq \frac{π}{4}

a \leq u \leq b

then

a \leq u and u \leq b

0 \leq arctan (x) and arctan (x) \leq \frac{π}{4}

0 \leq arctan (x) : x \geq 0

0 \leq arctan (x)

Switch sides

arctan (x) \geq 0

arctan (x) \geq a

then

x \geq tan (a)

x \geq 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

x \geq 0

arctan (x) \leq \frac{π}{4} : x \leq 1

arctan (x) \leq \frac{π}{4}

arctan (x) \leq a

then

x \leq tan (a)

x \leq tan (\frac{π}{4})

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

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}

= 1

x \leq 1

Combine the intervals

x \geq 0 and x \leq 1

Merge Overlapping Intervals

x \geq 0 and x \leq 1

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

x \geq 0

and

x \leq 1

0 \leq x \leq 1

0 \leq x \leq 1

sin (θ) < 0 and tan (θ) > 0

- 1 < sin^{2} (x) < 1

cot (θ) = \frac{21}{20} and cos (θ) > 0

sin (θ) = \frac{3}{2} and tan (θ) > 0

sin (θ) = - \frac{6}{9} and tan (θ) > 0