trigonometric inequalities

junaidnawaz

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

[tex] |\gamma| \leq \cos (\beta) [/tex]
[tex] \arccos (|\gamma|) \leq \beta [/tex]

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

tiny-tim

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,
[tex] 0 \leq \beta \leq \pi /2 [/tex]
[tex]-1 \leq \gamma \leq +1 [/tex]

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

tiny-tim

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
[tex] |\gamma| \leq \cos( \beta ) [/tex]

then

[tex] x \leq \beta[/tex]

can i find "x", by keeping the RHS fixed to [tex] \beta[/tex]

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

[tex] |\gamma| \leq \cos ( \beta ) [/tex]

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

[tex] \arccos( |\gamma| ) \geq \beta [/tex]

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

[tex] x \leq \beta [/tex]

what would be x ??

junaidnawaz

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

tiny-tim

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

junaidnawaz

Thank you :)