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

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Discussion Overview

The discussion revolves around the implications of applying the arccosine function to both sides of an inequality, specifically examining whether this operation preserves the relational operator. The context includes mathematical reasoning and exploration of inequalities involving trigonometric functions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether applying arccos() to both sides of the inequality |\gamma| ≤ cos(β) changes the relational operator.
  • Another participant notes that arccos is defined on the interval [0, π) and states that if β is also in this interval, the equations remain equivalent due to the monotonicity of cos and arccos.
  • A participant specifies the range of parameters as 0 ≤ β ≤ π/2 and -1 ≤ γ ≤ +1, seeking clarification on whether the relational operator changes from ≤ to ≥ when applying arccos().
  • It is suggested that since cos is decreasing, the operator does change from ≤ to ≥ when applying arccos().
  • Another participant asks if it is possible to find a variable x such that x ≤ β while keeping the right-hand side fixed to β, given the original inequality |\gamma| ≤ cos(β).
  • A participant expresses confusion about the nature of x, questioning if it is simply |γ|.
  • One participant asserts that it is not possible to maintain the operator as ≤ while keeping β on the right side unless β is replaced by a decreasing function of β.

Areas of Agreement / Disagreement

Participants express differing views on whether applying arccos() preserves the relational operator, with some asserting it changes while others explore the implications of keeping the operator the same. The discussion remains unresolved regarding the conditions under which the operator may or may not change.

Contextual Notes

Participants discuss specific ranges for β and γ, which may influence the behavior of the functions involved. The implications of monotonicity and the nature of the functions are central to the discussion.

junaidnawaz
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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??
 
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welcome to pf!

hi junaidnawaz! welcome to pf! :smile:

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

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

oh, i missed that! :rolleyes:

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

(but, eg, sin is increasing, so the ≤ would stay the same :wink:)
 
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 ??
 
sorry, i don't understand …

isn't x just |γ| ? :confused:
 
[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 ??
 
I wounder if its not a stupid question ... :P
 
junaidnawaz said:
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]

that's not possible (unless you replace β by some decreasing function of β, such as 1/β or -β)
 
  • #10
Thank you :)
 

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