# Trigonometric inequalities

 P: 6 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??
 Sci Advisor HW Helper Thanks P: 26,148 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)
 P: 6 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 ??
HW Helper
Thanks
P: 26,148
Trigonometric inequalities

 Quote by junaidnawaz 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 )
 P: 6 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 ??
 Sci Advisor HW Helper Thanks P: 26,148 sorry, i don't understand … isn't x just |γ| ?
 P: 6 $$|\gamma| \leq \cos ( \beta )$$ 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 ??
 P: 6 I wounder if its not a stupid question .... :P
 Quote by junaidnawaz however, i want to keep $$\beta$$ on right side, and i want to keep the relational operator as $$\leq$$, i.e., $$x \leq \beta$$