How can I prove sin2(x)≠sin⁡(x)

Question

Karsyn Wu · Accepted Answer

A lot of people like to consider the square root

f : x \mapsto \sqrt{x}

as the inverse function of the square

g : x \mapsto x^{2}

. This is true if you consider g as a function on

I = [0, \infty)

. Then you get

f \circ g (x) = x = g \circ f (x)

for all

x \in I

.
Nevertheless, you can consider the composition

f \circ g

not just on I but on

R

and you get

f \circ g (x) = | x |

, therefore

f \circ g

is not the identity on

R

. For a function h we can conclude

f \circ g \circ h = h

is true if and only if

h (x) \geq 0

for all x, because if there exists

x_{0}

such that

h (x_{0}) < 0

, then

f \circ g \circ h (x_{0}) = | h (x_{0}) | = - h (x_{0}) > 0 > h (x_{0})

.
And that is the case for

\sin

on

R

since

\sin (\frac{3}{2} π) = - 1 < 0

and therefore

\sqrt{\sin^{2} (\frac{32}{π})} = | \sin (\frac{32}{π}) | = 1 \neq - 1 = \sin (\frac{32}{π})

.
Otherwise you can consider

J = [0, π]

restricted on

\sin (x) \geq 0

and you get

x \in J

for all

x \in J

and then

\sqrt{\sin^{2} (x)} = | \sin (x) | = \sin (x)

.
You see that the statement depends on the domain of x.

How can I prove \(\displaystyle\sqrt{{{{\sin}^{{2}}{\left({x}\right)}}}}\ne{\sin{{\left({x}\right)}}}\)