2022-01-27

Find the least value for $\mathrm{sin}x-{\mathrm{cos}}^{2}x-1$
At first I found the first derivative to be ${y}^{\prime }=\mathrm{cos}x+2\mathrm{sin}x\mathrm{cos}x$
Critical point $0,-\frac{\pi }{6}$ (principal)
$y-\mathrm{sin}x+2\left(\mathrm{cos}2x\right)$
Then substitution of x by critical points I found minima. But my answer is incorrect.
Correct minimum value is $-\frac{9}{4}$

Aiden Cooper

Expert

$f\left(x\right)={\mathrm{sin}}^{2}x+\mathrm{sin}\left\{x\right\}-2={\left(\mathrm{sin}\left\{x\right\}+\frac{1}{2}\right)}^{2}-2.25\ge -2.25$
The equality occurs for $\mathrm{sin}x=-\frac{12}{}$, which says that -2.25 is a minimal value.

Micah May

Expert

Yes your way is correct indeed for $x=-\frac{\pi }{6}$ we have
$f\left(x\right)=-\frac{12}{-}\frac{34}{-}1=-\frac{94}{}$
As an alternative, recall that ${\mathrm{cos}}^{2}x=1-{\mathrm{sin}}^{2}x$ and therefore we have
$f\left(x\right)=\mathrm{sin}x-{\mathrm{cos}}^{2}x-1={\mathrm{sin}}^{2}x+\mathrm{sin}x-2$
then set $t=\mathrm{sin}x$ and consider the parabola
$g\left(t\right)={t}^{2}+t-2⇒{g}^{\prime }\left(t\right)=2t+1=0⇒t=-\frac{1}{2}$

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