Obel1x wrote: $\star$ Find maximal value of the function $f(x)=8-3\sin^2 (3x)+6 \sin (6x)$ Using identities $\sin(2u)=2\sin u \cos u$ and $\cos u=\sqrt{1-\sin^2 u}$ we need to find a maximum of a function: $f(x)=8-3\sin^2 (3x)+12 \sin (3x) \sqrt{1-\sin^2 (3x)}$ Now using supstitution $\sin(3x)=t$, we need to find maximum of a function: $g(t)=8-3t^2+12t\sqrt{1-t^2}$. Denote that maximum with $M$ so this inequality is equivalent to: $g(t)=M-(A\cdot t -B\cdot\sqrt{1-t^2})^2$ for $A,B$ which satisfy $A^2-B^2=3$, and $AB=6$ (and $M=8+B^2$) Solving this system of equations we get: $B^2=\frac{3}{2}(\sqrt{17}-1)$ so $\boxed{ M=\frac{1}{2}(13+3\sqrt{17}) }$

Obel1x wrote: $\star$ Find maximal value of the function $f(x)=8-3\sin^2 (3x)+6 \sin (6x)$ without calculus we know $\mid{acos\theta+bsin\theta}\mid\leq{\sqrt{a^2+b^2}}$ $f(x)=8-3\sin^2 (3x)+6 \sin (6x)=\frac{3}{2}cos6x+6sin6x+\frac{13}{2}\leq{\sqrt{36+\frac{9}{4}}+\frac{13}{2}}$$=\frac{13+3\sqrt{17}}{2}$

mcrasher wrote: Obel1x wrote: $\mid{acos\theta+bsin\theta}\mid\leq{\sqrt{a^2+b^2}}$ what is that?

mcrasher wrote: Obel1x wrote: $\star$ Find maximal value of the function $f(x)=8-3\sin^2 (3x)+6 \sin (6x)$ without calculus we know $\mid{acos\theta+bsin\theta}\mid\leq{\sqrt{a^2+b^2}}$ $f(x)=8-3\sin^2 (3x)+6 \sin (6x)=\frac{3}{2}cos6x+6sin6x+\frac{13}{2}\leq{\sqrt{36+\frac{9}{4}}+\frac{13}{2}}$$=\frac{13+3\sqrt{17}}{2}$ Yes, this it is the best solution!

reeh_haan wrote: mcrasher wrote: Obel1x wrote: $\mid{acos\theta+bsin\theta}\mid\leq{\sqrt{a^2+b^2}}$ what is that? This is Cauchy-Schwarz!