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Trigonometric inequalities 
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#1
Feb1111, 03:29 AM

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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?? 


#2
Feb1111, 03:42 AM

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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) 


#3
Feb1111, 04:08 AM

P: 6

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 bothsides, would it change the operator (from [tex] \leq[/tex] to [tex] \geq[/tex] ) or would it remain same ?? 


#4
Feb1111, 04:21 AM

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Trigonometric inequalities
yes, cos is decreasing, so the ≤ changes to ≥ (but, eg, sin is increasing, so the ≤ would stay the same ) 


#5
Feb1111, 05:08 AM

P: 6

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 ?? 


#6
Feb1111, 05:24 AM

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sorry, i don't understand …
isn't x just γ ? 


#7
Feb1111, 05:32 AM

P: 6

[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 ?? 


#8
Feb1111, 05:54 AM

P: 6

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



#9
Feb1111, 06:08 AM

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#10
Feb1111, 07:15 AM

P: 6

Thank you :)



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