Ahmed Stewart

2022-01-25

Proving $\frac{1+\mathrm{cos}x}{2+\mathrm{sin}x}<\frac{4}{3}$

Johnny Cummings

Expert

If $\mathrm{cos}\left(\theta \right)=\frac{3}{5}$ and $\mathrm{sin}\left(\theta \right)=\frac{4}{5}$ (which is possible since ${3}^{2}+{4}^{2}={5}^{2}$), the inequality (with $\le$ rather than <) is equivalent to $\mathrm{cos}\left(x-\theta \right)\le 1$

utgyrnr0

Expert

We want to prove $\frac{1+\mathrm{cos}x}{2+\mathrm{sin}x}<\frac{4}{3}$. By multiplying by $3\left(2+\mathrm{sin}x\right)$ (note that $2+\mathrm{sin}x>0$), we get that this is equivalent to $3\left(1+\mathrm{cos}x\right)\le 4\left(2+\mathrm{sin}x\right)$ which is equivalent to prove that $3\mathrm{cos}x-4\mathrm{sin}x\le 5$
This inequality comes from Cauchy Schwartz. Note that
${\left(3\mathrm{cos}x-4\mathrm{sin}x\right)}^{2}\le \left({3}^{2}+{\left(-4\right)}^{2}\right)\left({\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x\right)=25$

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