# Find the range of the function f(x)=cos(x)[sin(x)+sqrt(sin^2(x)+1/2)]

Find the range of the function $f\left(x\right)=\mathrm{cos}\left(x\right)\left[\mathrm{sin}\left(x\right)+\sqrt{{\mathrm{sin}}^{2}\left(x\right)+\frac{1}{2}}\right]$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

boursecasa2je
$f\left(x\right)-\mathrm{cos}x\mathrm{sin}x=\mathrm{cos}x\sqrt{\frac{1}{2}+{\mathrm{sin}}^{2}x}⇒f\left(x{\right)}^{2}-2f\left(x\right)\mathrm{cos}x\mathrm{sin}x=\frac{1}{2}{\mathrm{cos}}^{2}x⇒\phantom{\rule{0ex}{0ex}}4f\left(x\right)\mathrm{cos}x\mathrm{sin}x=2f\left(x{\right)}^{2}-{\mathrm{cos}}^{2}x⇒16f\left(x{\right)}^{2}{\mathrm{cos}}^{2}x\left(1-{\mathrm{cos}}^{2}x\right)=\left(2f\left(x{\right)}^{2}-{\mathrm{cos}}^{2}x{\right)}^{2}$

$16yt\left(1-t\right)=\left(2y-t{\right)}^{2}⇒4{y}^{2}+\left(16{t}^{2}-20t\right)y+{t}^{2}=0⇒\phantom{\rule{0ex}{0ex}}8y{y}^{\prime }+\left(16{t}^{2}-20t\right){y}^{\prime }+\left(32t-20\right)y+2t=0$
Here ${y}^{\prime }=\frac{dy}{dt}$. In critical points $\frac{df\left(x\right)}{dx}=0⇒\frac{dy}{dt}=0$
$\left(32t-20\right)y+2t=0⇒y=-\frac{t}{16t-10}⇒\frac{4{t}^{2}}{\left(16t-10{\right)}^{2}}-\frac{t\left(16{t}^{2}-20t\right)}{16t-10}+{t}^{2}=0⇒\phantom{\rule{0ex}{0ex}}4{t}^{2}-t\left(16{t}^{2}-20t\right)\left(16t-10\right)+{t}^{2}\left(16t-10{\right)}^{2}=0⇒32{t}^{2}\left(5t-3\right)=0$
$t=0$ gives minimum $y=0$. $t=\frac{3}{5}$ gives maximum $y=\frac{3}{2}$
Minimum $f\left(x\right)=-\sqrt{\frac{3}{2}}$ is obtained at $\mathrm{cos}x=-\sqrt{\frac{3}{5}}$, $\mathrm{sin}x=\sqrt{\frac{2}{5}}$
Maximum $f\left(x\right)=\sqrt{\frac{3}{2}}$ is obtained at $\mathrm{cos}x=\sqrt{\frac{3}{5}}$, $\mathrm{sin}x=\sqrt{\frac{2}{5}}$