parity csc(csc(x)) — step-by-step solution

Let's study the parity of

\csc (\csc (x))

with respect to

x

The cosecant is read as

\csc (u) = \frac{1}{\sin (u)}

, on each interval where

\sin (u) \neq 0

The domain is

{x \neq k π, k \in ℤ, \sin (\frac{1}{\sin (x)}) \neq 0}

Replacing

x

by -

x

does not change the conditions of the domain:

\sin (\frac{1}{\sin (- x)}) = - \sin (\frac{1}{\sin (x)})

, and

- \sin (\frac{1}{\sin (x)}) \neq 0

is equivalent to

\sin (\frac{1}{\sin (x)}) \neq 0

This domain is symmetric about 0: if

x

is in the domain, -

x

is too

Replace

x

by -

x

: f(-

x

) =

\frac{1}{\sin (\frac{1}{\sin (- x)})}

⇥

\frac{1}{\sin (\frac{1}{\sin (- x)})}

⇥ⓘ sin is odd, sin(-x) = -sin(x):

\sin (- x)

- \sin (x)

⇥

\frac{1}{\sin (\frac{1}{- \sin (x)})}

⇥ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{1}{- \sin (x)}

- \frac{1}{\sin (x)}

■sin is odd, sin(-x) = -sin(x):

\sin (- \frac{1}{\sin (x)})

- \sin (\frac{1}{\sin (x)})

⇥

\frac{1}{- \sin (\frac{1}{\sin (x)})}

⇥ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{1}{- \sin (\frac{1}{\sin (x)})}

- \frac{1}{\sin (\frac{1}{\sin (x)})}

⇥

- \frac{1}{\sin (\frac{1}{\sin (x)})}

f(-

x

) = -f(

x

): the function is odd

▶Parity: $odd$