2.恒等变换（二）

$y=a\sin x+b\cos x=\sqrt{a^2+b^2}({\displaystyle \frac{a}{\sqrt{a^2+b^2}}}\sin x+{\displaystyle \frac{b}{\sqrt{a^2+b^2}}}\cos x)$
$=\sqrt{a^2+b^2}\sin (x+\phi )$
$\mathrm{这}\mathrm{里}\mathrm{设}\cos \phi ={\displaystyle \frac{a}{\sqrt{a^2+b^2}}},\sin \phi ={\displaystyle \frac{b}{\sqrt{a^2+b^2}}}$

$\mathrm{今}\mathrm{天}\mathrm{我}\mathrm{们}\mathrm{选}\mathrm{择}\mathrm{用}\mathrm{角}\mathrm{度}\mathrm{当}\mathrm{变}\mathrm{量}\mathrm{构}\mathrm{建}\mathrm{函}\mathrm{数}$
$Rt\mathrm{△}BOC\mathrm{中}，BC=OC\sin \alpha =\sin \alpha ,OB=OC\cos \alpha =\cos \alpha$
$AB=OB-OA$
$Rt\mathrm{△}AOD\mathrm{中}，$
${\displaystyle \frac{AD}{OA}}={\displaystyle \frac{BC}{OA}}=\tan {\displaystyle \frac{\pi }{3}}$
$\therefore OA={\displaystyle \frac{\sqrt{3}}{3}}\sin \alpha ,AB=\cos \alpha -{\displaystyle \frac{\sqrt{3}}{3}}\sin \alpha$
$\mathrm{设}\mathrm{矩}\mathrm{形}\mathrm{的}\mathrm{面}\mathrm{积}S$
$S=AB\cdot BC=(\cos \alpha -{\displaystyle \frac{\sqrt{3}}{3}}\sin \alpha )\sin \alpha$
$=\sin \alpha \cos \alpha -{\displaystyle \frac{\sqrt{3}}{3}}{\sin }^2\alpha$
$\mathrm{这}\mathrm{是}\mathrm{我}\mathrm{们}\mathrm{以}\mathrm{后}\mathrm{非}\mathrm{常}\mathrm{常}\mathrm{见}\mathrm{的}\mathrm{一}\mathrm{种}\mathrm{形}\mathrm{式}$
$\mathrm{降}\mathrm{次}\mathrm{化}\mathrm{一}(\mathrm{三}\mathrm{角}\mathrm{函}\mathrm{数}\mathrm{流}\mathrm{行}\mathrm{词}\mathrm{语})$
$S={\displaystyle \frac{1}{2}}\sin 2\alpha -{\displaystyle \frac{\sqrt{3}}{3}}({\displaystyle \frac{1-\cos 2\alpha }{2}})$
$={\displaystyle \frac{1}{2}}\sin 2\alpha +{\displaystyle \frac{\sqrt{3}}{6}}\cos 2\alpha -{\displaystyle \frac{\sqrt{3}}{3}}$
($\mathrm{提}\mathrm{取}\sqrt{({\displaystyle \frac{1}{2}}{)}^2+({\displaystyle \frac{\sqrt{3}}{6}}{)}^2}={\displaystyle \frac{1}{\sqrt{3}}}$)
(中间过程不必须有理化，减少不必要计算)
$S={\displaystyle \frac{1}{\sqrt{3}}}({\displaystyle \frac{\sqrt{3}}{2}}\sin x+{\displaystyle \frac{1}{2}}\cos x)-{\displaystyle \frac{\sqrt{3}}{3}}$
$={\displaystyle \frac{1}{\sqrt{3}}}\sin (2\alpha +{\displaystyle \frac{\pi }{6}})-{\displaystyle \frac{\sqrt{3}}{3}}(0<\alpha <{\displaystyle \frac{\pi }{3}})$
$\alpha \mathrm{角}\mathrm{的}\mathrm{范}\mathrm{围}\mathrm{中}\mathrm{没}\mathrm{有}\mathrm{等}\mathrm{号}？$
${\displaystyle \frac{\pi }{6}}<2\alpha +{\displaystyle \frac{\pi }{6}}<{\displaystyle \frac{2\pi }{3}}$
$\therefore \mathrm{当}2\alpha +{\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\pi }{2}},\mathrm{即}\alpha ={\displaystyle \frac{\pi }{6}}\mathrm{时}，S\mathrm{的}\mathrm{最}\mathrm{大}\mathrm{值}\mathrm{为}{\displaystyle \frac{\sqrt{3}}{6}}$

$\mathrm{设}BC=x$
$Rt\mathrm{△}BOC\mathrm{中}，OB=\sqrt{1-x^2}$
$Rt\mathrm{△}AOD\mathrm{中}，OA={\displaystyle \frac{\sqrt{3}}{3}}x$
$AB=OB-OA=\sqrt{1-x^2}-{\displaystyle \frac{\sqrt{3}}{3}}x$
$S=x(\sqrt{1-x^2}-{\displaystyle \frac{\sqrt{3}}{3}}x)(0<x<{\displaystyle \frac{\sqrt{3}}{2}})$
会陷入困境
其实
$\mathrm{可}\mathrm{以}\mathrm{采}\mathrm{取}\mathrm{三}\mathrm{角}\mathrm{换}\mathrm{元}，\mathrm{令}x=\sin \theta (0<\theta <{\displaystyle \frac{\pi }{3}})$
$S=\sin \theta (\cos \theta -{\displaystyle \frac{\sqrt{3}}{3}}\sin \theta )$

$\mathrm{令}\sin x+\cos x=t\in [-\sqrt{2},\sqrt{2}]$
$\mathrm{根}\mathrm{据}\mathrm{三}\mathrm{角}\mathrm{基}\mathrm{本}\mathrm{关}\mathrm{系}\mathrm{式}，\sin 2x=(\sin x+\cos x{)}^2-1$
$\therefore y=t^2+t-1=(t+{\displaystyle \frac{1}{2}}{)}^2-{\displaystyle \frac{5}{4}},t\in [-\sqrt{2},\sqrt{2}]$
$y_{max}=1+\sqrt{2},y_{min}=-{\displaystyle \frac{5}{4}}$