Does applying arccos() to both sides of an inequality preserve its relation?

[junaidnawaz]

Please help me to confirm, weather the following step is correct

|\gamma| \leq \cos (\beta)
\arccos (|\gamma|) \leq \beta

does taking the arccos() on both sides of equation changes the relational operator??

 

[tiny-tim]

welcome to pf!

hi junaidnawaz! welcome to pf!

(have a beta: β and a gamma: γ and a ≤ )

arccos is defined as being in [0,π)

so long as β is also in [0,π), your equations are the same (because cos is monotone in that region, and therefore so is arccos)

 

[junaidnawaz]

Thx v much for your reply.

in my case, the range of parameters is as,
0 \leq \beta \leq \pi /2
-1 \leq \gamma \leq +1

by taking arccos() on both-sides, would it change the operator (from \leq to \geq ) or would it remain same ??

 

[tiny-tim]

by taking arccos() on both-sides, would it change the operator (from \leq to \geq ) or would it remain same ??

oh, i missed that!

yes, cos is decreasing, so the ≤ changes to ≥

(but, eg, sin is increasing, so the ≤ would stay the same )

 

[junaidnawaz]

Thank you.

if
|\gamma| \leq \cos( \beta )

then

x \leq \beta

can i find "x", by keeping the RHS fixed to \beta

is this possible to find x ?? by keeping RHS and relation operator the same ??

 

[tiny-tim]

sorry, i don't understand …

isn't x just |γ| ?

 

[junaidnawaz]

|\gamma| \leq \cos ( \beta )

When I take arccos() on both sides, it becomes

\arccos( |\gamma| ) \geq \beta

however, i want to keep \beta on right side, and i want to keep the relational operator as \leq, i.e.,

x \leq \beta

what would be x ??

 

[junaidnawaz]

I wounder if its not a stupid question ... :P

 

[tiny-tim]

however, i want to keep \beta on right side, and i want to keep the relational operator as \leq, i.e.,

x \leq \beta

that's not possible (unless you replace β by some decreasing function of β, such as 1/β or -β)

 

[junaidnawaz]

Thank you :)