2025年上學(xué)期高一數(shù)學(xué)變式題訓(xùn)練（二）一、三角函數(shù)基礎(chǔ)運(yùn)算與變式題型1：同角三角函數(shù)關(guān)系應(yīng)用典例1已知$\sin\alpha=\frac{3}{5}$，且$\alpha$為第二象限角，求$\cos\alpha$和$\tan\alpha$的值。解析由平方關(guān)系$\sin^2\alpha+\cos^2\alpha=1$得：$\cos^2\alpha=1-\left(\frac{3}{5}\right)^2=\frac{16}{25}$∵$\alpha$為第二象限角，$\cos\alpha<0$，∴$\cos\alpha=-\frac{4}{5}$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-\frac{3}{4}$變式1-1已知$\tan\theta=2$，求$\frac{\sin\theta+2\cos\theta}{3\sin\theta-\cos\theta}$的值。解析分子分母同除以$\cos\theta$（$\cos\theta\neq0$）：原式$=\frac{\tan\theta+2}{3\tan\theta-1}=\frac{2+2}{6-1}=\frac{4}{5}$變式1-2已知$\sin\alpha+\cos\alpha=\frac{1}{5}$，$\alpha\in(0,\pi)$，求$\sin\alpha-\cos\alpha$的值。解析平方得$(\sin\alpha+\cos\alpha)^2=\frac{1}{25}\Rightarrow1+2\sin\alpha\cos\alpha=\frac{1}{25}\Rightarrow\sin\alpha\cos\alpha=-\frac{12}{25}$∵$\alpha\in(0,\pi)$，$\sin\alpha>0$，$\cos\alpha<0$，∴$\sin\alpha-\cos\alpha>0$$(\sin\alpha-\cos\alpha)^2=1-2\sin\alpha\cos\alpha=1+\frac{24}{25}=\frac{49}{25}\Rightarrow\sin\alpha-\cos\alpha=\frac{7}{5}$題型2：誘導(dǎo)公式與恒等變換典例2化簡：$\sin(2\pi-\alpha)\cos(\pi+\alpha)\tan(\pi-\alpha)$解析原式$=(-\sin\alpha)(-\cos\alpha)(-\tan\alpha)=-\sin\alpha\cos\alpha\cdot\frac{\sin\alpha}{\cos\alpha}=-\sin^2\alpha$變式2-1計算：$\sin\frac{25\pi}{6}+\cos\frac{25\pi}{3}+\tan\left(-\frac{25\pi}{4}\right)$解析$\sin\frac{25\pi}{6}=\sin\left(4\pi+\frac{\pi}{6}\right)=\sin\frac{\pi}{6}=\frac{1}{2}$$\cos\frac{25\pi}{3}=\cos\left(8\pi+\frac{\pi}{3}\right)=\cos\frac{\pi}{3}=\frac{1}{2}$$\tan\left(-\frac{25\pi}{4}\right)=-\tan\left(6\pi+\frac{\pi}{4}\right)=-\tan\frac{\pi}{4}=-1$原式$=\frac{1}{2}+\frac{1}{2}-1=0$變式2-2已知$\cos(\pi+\alpha)=\frac{3}{5}$，$\alpha\in\left(\frac{\pi}{2},\pi\right)$，求$\sin\left(\alpha-\frac{3\pi}{2}\right)$的值。解析$\cos(\pi+\alpha)=-\cos\alpha=\frac{3}{5}\Rightarrow\cos\alpha=-\frac{3}{5}$$\sin\left(\alpha-\frac{3\pi}{2}\right)=\sin\alpha\cos\frac{3\pi}{2}-\cos\alpha\sin\frac{3\pi}{2}=\cos\alpha=-\frac{3}{5}$題型3：三角函數(shù)圖像與性質(zhì)典例3函數(shù)$f(x)=A\sin(\omegax+\varphi)(A>0,\omega>0,|\varphi|<\frac{\pi}{2})$的部分圖像如圖所示，求其解析式。解析由圖像知$A=2$，$T=\frac{2\pi}{\omega}=4\times\left(\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right)=2\pi\Rightarrow\omega=1$將點$\left(\frac{\pi}{3},2\right)$代入：$2\sin\left(\frac{\pi}{3}+\varphi\right)=2\Rightarrow\frac{\pi}{3}+\varphi=\frac{\pi}{2}+2k\pi$∵$|\varphi|<\frac{\pi}{2}$，∴$\varphi=\frac{\pi}{6}$，故$f(x)=2\sin\left(x+\frac{\pi}{6}\right)$變式3-1求函數(shù)$y=2\sin\left(2x-\frac{\pi}{3}\right)+1$的單調(diào)遞增區(qū)間。解析令$-\frac{\pi}{2}+2k\pi\leq2x-\frac{\pi}{3}\leq\frac{\pi}{2}+2k\pi$解得$-\frac{\pi}{12}+k\pi\leqx\leq\frac{5\pi}{12}+k\pi$，$k\in\mathbb{Z}$遞增區(qū)間為$\left-\frac{\pi}{12}+k\pi,\frac{5\pi}{12}+k\pi\right$變式3-2函數(shù)$f(x)=\sinx+\cosx$在區(qū)間$[0,\pi]$上的最大值和最小值。解析$f(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$，$x\in[0,\pi]\Rightarrowx+\frac{\pi}{4}\in\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$當(dāng)$x+\frac{\pi}{4}=\frac{\pi}{2}$即$x=\frac{\pi}{4}$時，$f(x){\text{max}}=\sqrt{2}$當(dāng)$x+\frac{\pi}{4}=\frac{5\pi}{4}$即$x=\pi$時，$f(x){\text{min}}=-1$二、平面向量線性運(yùn)算與數(shù)量積題型4：向量線性表示典例4在$\triangleABC$中，$D$為$BC$中點，$\overrightarrow{AB}=\vec{a}$，$\overrightarrow{AC}=\vec$，用$\vec{a},\vec$表示$\overrightarrow{AD}$。解析$\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\vec{a}+\frac{1}{2}\overrightarrow{BC}=\vec{a}+\frac{1}{2}(\vec-\vec{a})=\frac{1}{2}\vec{a}+\frac{1}{2}\vec$變式4-1在$\triangleABC$中，$\overrightarrow{AN}=\frac{1}{3}\overrightarrow{NC}$，$P$是$BN$上一點，若$\overrightarrow{AP}=m\overrightarrow{AB}+\frac{2}{11}\overrightarrow{AC}$，求$m$的值。解析設(shè)$\overrightarrow{BP}=k\overrightarrow{BN}$，則$\overrightarrow{AP}=\overrightarrow{AB}+k\overrightarrow{BN}=\vec{a}+k\left(\overrightarrow{AN}-\overrightarrow{AB}\right)=\vec{a}+k\left(\frac{1}{4}\vec-\vec{a}\right)=(1-k)\vec{a}+\frac{k}{4}\vec$對比$\overrightarrow{AP}=m\vec{a}+\frac{2}{11}\vec$，得$\frac{k}{4}=\frac{2}{11}\Rightarrowk=\frac{8}{11}$，$m=1-\frac{8}{11}=\frac{3}{11}$題型5：向量坐標(biāo)運(yùn)算典例5已知$\vec{a}=(2,1)$，$\vec=(-1,3)$，求$2\vec{a}-3\vec$，$\vec{a}\cdot\vec$，$|\vec{a}+\vec|$。解析$2\vec{a}-3\vec=2(2,1)-3(-1,3)=(4+3,2-9)=(7,-7)$$\vec{a}\cdot\vec=2\times(-1)+1\times3=1$$\vec{a}+\vec=(1,4)$，$|\vec{a}+\vec|=\sqrt{1^2+4^2}=\sqrt{17}$變式5-1已知$\vec{a}=(1,\sqrt{3})$，$\vec=(\sqrt{3},1)$，求$\vec{a}$與$\vec$的夾角$\theta$。解析$\cos\theta=\frac{\vec{a}\cdot\vec}{|\vec{a}||\vec|}=\frac{1\times\sqrt{3}+\sqrt{3}\times1}{2\times2}=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$∵$\theta\in[0,\pi]$，∴$\theta=\frac{\pi}{6}$變式5-2已知$\vec{a}=(m,2)$，$\vec=(3,-1)$，若$\vec{a}\perp\vec$，求$m$；若$\vec{a}\parallel\vec$，求$m$。解析$\vec{a}\perp\vec\Rightarrow\vec{a}\cdot\vec=3m-2=0\Rightarrowm=\frac{2}{3}$$\vec{a}\parallel\vec\Rightarrow-m-6=0\Rightarrowm=-6$題型6：向量數(shù)量積應(yīng)用典例6在$\triangleABC$中，$\overrightarrow{AB}\cdot\overrightarrow{AC}=|\overrightarrow{AB}-\overrightarrow{AC}|=2$，求$\triangleABC$面積的最大值。解析設(shè)$|\overrightarrow{AB}|=c$，$|\overrightarrow{AC}|=b$，則$\overrightarrow{AB}\cdot\overrightarrow{AC}=bc\cosA=2$$|\overrightarrow{AB}-\overrightarrow{AC}|=|\overrightarrow{BC}|=a=2$，由余弦定理$a^2=b^2+c^2-2bc\cosA\Rightarrow4=b^2+c^2-4\Rightarrowb^2+c^2=8$∵$b^2+c^2\geq2bc\Rightarrowbc\leq4$，當(dāng)且僅當(dāng)$b=c=2$時取等號$\cosA=\frac{2}{bc}\geq\frac{1}{2}\RightarrowA\in\left(0,\frac{\pi}{3}\right]$，$\sinA=\sqrt{1-\cos^2A}=\sqrt{1-\frac{4}{b^2c^2}}$$S_{\triangleABC}=\frac{1}{2}bc\sinA=\frac{1}{2}\sqrt{b^2c^2-4}\leq\frac{1}{2}\sqrt{16-4}=\sqrt{3}$變式6-1已知$\vec{a}$，$\vec$是單位向量，且$\vec{a}+\vec=(1,1)$，求$|\vec{a}-\vec|$。解析$|\vec{a}+\vec|^2=(\vec{a}+\vec)^2=|\vec{a}|^2+2\vec{a}\cdot\vec+|\vec|^2=2+2\vec{a}\cdot\vec=2\Rightarrow\vec{a}\cdot\vec=0$$|\vec{a}-\vec|^2=|\vec{a}|^2-2\vec{a}\cdot\vec+|\vec|^2=2\Rightarrow|\vec{a}-\vec|=\sqrt{2}$三、綜合應(yīng)用與拓展題型7：三角函數(shù)與向量綜合典例7已知向量$\vec{m}=(\sinx,\cosx)$，$\vec{n}=(\cosx,\sqrt{3}\cosx)$，函數(shù)$f(x)=\vec{m}\cdot\vec{n}-\frac{\sqrt{3}}{2}$。（1）求$f(x)$的最小正周期；（2）求$f(x)$在$[0,\frac{\pi}{2}]$上的單調(diào)區(qū)間。解析（1）$f(x)=\sinx\cosx+\sqrt{3}\cos^2x-\frac{\sqrt{3}}{2}=\frac{1}{2}\sin2x+\frac{\sqrt{3}}{2}(1+\cos2x)-\frac{\sqrt{3}}{2}=\frac{1}{2}\sin2x+\frac{\sqrt{3}}{2}\cos2x=\sin\left(2x+\frac{\pi}{3}\right)$$T=\frac{2\pi}{2}=\pi$（2）令$-\frac{\pi}{2}+2k\pi\leq2x+\frac{\pi}{3}\leq\frac{\pi}{2}+2k\pi\Rightarrow-\frac{5\pi}{12}+k\pi\leqx\leq\frac{\pi}{12}+k\pi$在$[0,\frac{\pi}{2}]$上，增區(qū)間為$[0,\frac{\pi}{12}]$，減區(qū)間為$[\frac{\pi}{12},\frac{\pi}{2}]$變式7-1已知$\vec{a}=(\sin\theta,1)$，$\vec=(1,\cos\theta)$，$-\frac{\pi}{2}<\theta<\frac{\pi}{2}$，求$|\vec{a}+\vec|$的最大值。解析$\vec{a}+\vec=(\sin\theta+1,1+\cos\theta)$$|\vec{a}+\vec|^2=(\sin\theta+1)^2+(1+\cos\theta)^2=3+2(\sin\theta+\cos\theta)=3+2\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)$∵$\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$，$\theta+\frac{\pi}{4}\in\left(-\frac{\pi}{4},\frac{3\pi}{4}\right)$，$\sin\left(\theta+\frac{\pi}{4}\right)\leq1$∴$|\vec{a}+\vec|^2\leq3+2\sqrt{2}=(\sqrt{2}+1)^2\Rightarrow|\vec{a}+\vec|_{\text{max}}=\sqrt{2}+1$題型8：實際應(yīng)用問題典例8如圖，某摩天輪半徑為50米，中心距地面100米，逆時針方向勻速旋轉(zhuǎn)，旋轉(zhuǎn)一周需30分鐘。某人從摩天輪最低點$A$處登上摩天輪，經(jīng)過$t$分鐘后到達(dá)點$P$，設(shè)點$P$的縱坐標(biāo)為$y$（米），求$y$關(guān)于$t$的函數(shù)解析式，并求$t=7.5$分鐘時點$P$距離地面的高度。解析設(shè)$y=A\sin(\omegat+\varphi)+k$，周期$T=30\Rightarrow\omega=\frac{2\pi}{30}=\frac{\pi}{15}$振幅$A=50$，$k=100$，初始位置$t=0$時$y=50$，即$50=50\sin\varphi+100\Rightarrow\sin\varphi=-1\Rightarrow\varphi=-\frac{\pi}{2}$∴$y=50\sin\left(\frac{\pi}{15}t-\frac{\pi}{2}\right)+100=