\(\fbox{例30}\)(2017\(\cdot\)广东汕头一模)(求面积的最大值)
已知\(\Delta ABC\)的内角\(A,B,C\)的对边分别是\(a,b,c\),且满足\(b=c\),\(\cfrac{b}{a}=\cfrac{1-cosB}{cosA}\),若点\(O\)是\(\Delta ABC\)外的一点,\(\angle AOB=\theta(0<\theta<\pi)\),\(OA=2\),\(OB=1\),则四边形\(OACB\)面积的最大值是【】
\(A.\cfrac{4+5\sqrt{3}}{4} \;\; \;\;\) \(B.\cfrac{8+5\sqrt{3}}{4} \;\; \;\;\) \(C.3 \;\; \;\;\) \(D.\cfrac{4+5\sqrt{3}}{4}\)
分析:由\(\cfrac{b}{a}=\cfrac{sinB}{sinA}=\cfrac{1-cosB}{cosA}\),
得到\(sinBcosA+cosBsinA=sinA\),即\(sin(A+B)=sinA\)
则\(sinC=sinA\),即\(A=C\),
故\(a=b=c\),为等边三角形。
在\(\Delta AOB\)中,\(AB^2=2^2+1^2-2\cdot 2\cdot 1\cdot cos\theta=5-4cos\theta\),
故\(S_{OACB}=S_{\Delta AOB}+S_{\Delta ABC}\)
\(=\cfrac{1}{2}\cdot 2\cdot 1\cdot sin\theta+\cfrac{\sqrt{3}}{4}\cdot AB^2\)
\(=sin\theta+\cfrac{\sqrt{3}}{4}(5-4cos\theta)=2sin(\theta-\cfrac{\pi}{3})+\cfrac{5\sqrt{3}}{4}\)
当\(\theta-\cfrac{\pi}{3}=\cfrac{\pi}{2}\)时,即\(\theta=\cfrac{5\pi}{6}\in (0,\pi)\)时,四边形的面积有最大值,
且$S_{max}=2+\cfrac{5\sqrt{3}}{4}=\cfrac{8+5\sqrt{3}}{4} $,故选B。
\(\fbox{例30}\)(2017\(\cdot\)皖北协作区3月联考)(求取值范围)
如图,\(\angle BAC=\cfrac{2\pi}{3}\),\(P\)为\(\angle BAC\)内部一点,过点\(P\)的直线与\(\angle BAC\)的两边交于点\(B、C\),且\(PA\perp AC\),\(AP=\sqrt{3}\)。
(1)若\(AB=3\),求\(PC\);
分析:在\(\Delta ABP\)中,\(\angle BAP=30^{\circ}\),\(AB=3\),\(AP=\sqrt{3}\),
由余弦定理得到\(BP=\sqrt{3}\),故\(\angle BAP=\angle PBA=30^{\circ}\),
则\(\angle APC=60^{\circ}\),在\(Rt\Delta APC\)中,可得\(PC=2\sqrt{3}\)。
(2)求\(\cfrac{1}{PB}+\cfrac{1}{PC}\)的取值范围。
分析:设\(\angle PBA=\theta\),则\(\theta\in (0,\cfrac{\pi}{3})\);
(说明:当过点\(P\)的直线和\(AB\)平行时,\(\theta=0\);当过点\(P\)的直线和\(AC\)平行时,\(\theta=\cfrac{\pi}{3}\))
在\(\Delta ABP\)中,\(\angle BAP=30^{\circ}\),\(\angle PBA=\theta\),\(AP=\sqrt{3}\),
故由正弦定理得到\(\cfrac{PB}{sin30^{\circ}}=\cfrac{\sqrt{3}}{sin\theta}\),即\(PB=\cfrac{\cfrac{\sqrt{3}}{2}}{sin\theta}\);
在\(Rt\Delta APC\)中,\(\angle CPA=\theta+\cfrac{\pi}{6}\),\(PC=\cfrac{\sqrt{3}}{cos(\theta+\cfrac{\pi}{6})}\),
故\(\cfrac{1}{PB}+\cfrac{1}{PC}=\cfrac{2sin\theta}{\sqrt{3}}+\cfrac{cos(\theta+\cfrac{\pi}{6})}{\sqrt{3}}\)
\(=\cfrac{1}{\sqrt{3}}(\cfrac{3}{2}sin\theta+\cfrac{\sqrt{3}}{2}cos\theta)\)
\(=sin(\theta+\cfrac{\pi}{6})\),且\(\theta\in (0,\cfrac{\pi}{3})\);
故\(\cfrac{1}{2}<sin(\theta+\cfrac{\pi}{6})<1\)。
\(\fbox{例31}\)
函数\(f(x)=2cos(\omega x+\phi)(\omega\neq 0)\)对任意\(x\)都有\(f(\cfrac{\pi}{4}+x)=f(\cfrac{\pi}{4}-x)\)成立,则\(f(\cfrac{\pi}{4})\)的值为【】
A、\(2或0\) \(\hspace{2cm}\) B、\(-2或2\) \(\hspace{2cm}\) C、\(0\) \(\hspace{2cm}\) D、 \(-2或0\)
分析:由任意\(x\)都有\(f(\cfrac{\pi}{4}+x)=f(\cfrac{\pi}{4}-x)\)成立,
可知\(x=\cfrac{\pi}{4}\)为函数的一条对称轴,
而正弦型或余弦型函数在对称轴处必然会取到最值,
故\(f(\cfrac{\pi}{4})=\pm 2\),选B。
解后反思:此题目如果不注意函数的性质,往往会想到求\(\omega\)和\(\phi\),这样思路就跑偏了。
\(\fbox{例32}\)
已知\(\cfrac{\pi}{2}<\beta<\alpha<\cfrac{3\pi}{4}\),\(cos(\alpha-\beta)=\cfrac{12}{13}\),\(sin(\alpha+\beta)=-\cfrac{3}{5}\), 则\(sin\alpha+cos\alpha\)的值为_____。
分析:先由给定的不等式求解\(\alpha\pm \beta\)的范围,以便于求解其余名函数的值,为后续的计算打基础。
由\(\cfrac{\pi}{2}<\beta<\alpha<\cfrac{3\pi}{4}\),
得到\(\cfrac{\pi}{2}<\alpha<\cfrac{3\pi}{4}\),\(\cfrac{\pi}{2}<\beta<\cfrac{3\pi}{4}\),
故\(\pi<\alpha+\beta<\cfrac{3\pi}{2}\),由\(sin(\alpha+\beta)=-\cfrac{3}{5}\),
得到\(cos(\alpha+\beta)=-\cfrac{4}{5}\);
又由\(\cfrac{\pi}{2}<\alpha<\cfrac{3\pi}{4}\),\(-\cfrac{3\pi}{4}<-\beta<-\cfrac{\pi}{2}\),及\(\alpha-\beta>0\)
得到\(0<\alpha-\beta<\cfrac{\pi}{4}\),由\(cos(\alpha-\beta)=\cfrac{12}{13}\),
得到\(sin(\alpha-\beta)=\cfrac{5}{13}\);
\(sin2\alpha=sin[(\alpha+\beta)+(\alpha-\beta)]\)
\(=sin(\alpha+\beta)cos(\alpha-\beta)+cos(\alpha+\beta)sin(\alpha-\beta)=-\cfrac{56}{65}\)
又\((sin\alpha+cos\alpha)^2=1+sin2\alpha=\cfrac{9}{65}\)
又由于\(\cfrac{\pi}{2}<\beta<\alpha<\cfrac{3\pi}{4}\),借助三角函数线可知
则有\(sin\alpha+cos\alpha>0\),
故\(sin\alpha+cos\alpha=\sqrt{\cfrac{9}{65}}=\cfrac{3\sqrt{65}}{65}\)。
\(\fbox{例33}\)【2018高考一卷第16题】
求\(f(x)=2sinx+sin2x\)的最小值。
法1:\(f'(x)=2cosx+2cos2x=2cosx+2(2cos^x-1)\)
\(=4cos^2x+2cosx-2=(2cosx+2)(2cosx-1)\)
\(=4(cosx+1)(cosx-\cfrac{1}{2})\)
注意到\(cosx+1\ge 0\)恒成立,故
令\(f'(x)>0\)得到,\(cosx>\cfrac{1}{2}\),令\(f'(x)<0\)得到,\(cosx<\cfrac{1}{2}\),
则\(x\in [2k\pi-\cfrac{5\pi}{3},2k\pi-\cfrac{\pi}{3}](k\in Z)\)时,函数\(f(x)\)单调递减;
\(x\in [2k\pi-\cfrac{\pi}{3},2k\pi+\cfrac{\pi}{3}](k\in Z)\)时,函数\(f(x)\)单调递增;
故当\(x=2k\pi-\cfrac{\pi}{3}(k\in Z)\)时,\(f(x)_{min}=f(\cfrac{2k\pi-\pi}{3})=-\cfrac{3\sqrt{3}}{2}\)。
\(\fbox{例34}\)【2019届高三理科数学课时作业用题】
在斜三角形\(ABC\)中,\(sinA=-\sqrt{2}cosBcosC\),且\(tanBtanC=1-\sqrt{2}\),则角\(A\)的大小为【】
法1分析:由\(sinA=sin(B+C)=sinBcosC+cosBsinC=-\sqrt{2}cosBcosC\),
则\(\cfrac{sinBcosC+cosBsinC}{cosBcosC}=tanB+tanC=-\sqrt{2}\),
又由于\(tanBtanC=1-\sqrt{2}\),故\(-tanA=tan(B+C)=\cfrac{tanB+tanC}{1-tanBtanC}\),
即\(-tanA=\cfrac{-\sqrt{2}}{1-(1-\sqrt{2})}=-1\),故\(tanA=1\),又\(A\in (0,\pi)\),
则\(A=\cfrac{\pi}{4}\),故选\(A\)。