5.2.2 同角三角函数的基本关系
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【基础过关系列】2022-2023学年高一数学上学期同步知识点剖析精品讲义(人教A版2019)
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必修第一册同步巩固,难度2颗星!
基础知识
同角三角函数基本关系式
解释
由三角函数的定义,如下图,\(α\)的终边与单位圆交于点\(P(x,y)\),
可知\(\sin α=y\),\(\cos α=x\), \(\tan \alpha=\dfrac{y}{x}=(x \neq 0)\).
过点\(P\)作\(PH⊥x\)轴,在\(Rt∆PHO\)中,\(OH^2+PH^2=OP^2=1\),
因此\(x^2+y^2=1\),即\(\sin ^2 α+\cos ^2 α=1\),
当\(α≠π/2+kπ(k∈Z)\)时, \(\tan \alpha=\dfrac{y}{x}=\dfrac{\sin \alpha}{\cos \alpha}\).
拓展
\(\qquad\) \((\sin α+\cos α)^2=1+2 \sin α \cdot \cos α\);\(\qquad \qquad\) \((\sin α-\cos α)^2=1-2 \sin α \cdot \cos α\).
基本方法
【题型1】利用三角函数基本关系式求值
【典题1】 已知\(α∈(0,π)\),\(\tan α=-2\),则\(\cos α=\) \(\underline{\quad \quad}\).
解析 方法1 \(∵\tan α=-2\), \(\therefore \dfrac{\sin \alpha}{\cos \alpha}=-2\),即\(\sin α=-2\cos α\),
又 \(\sin ^2 \alpha+\cos ^2 \alpha=1 \Rightarrow \cos \alpha=\pm \dfrac{\sqrt{5}}{5}\),
\(∵α∈(0,π)\),且\(\tan α=-2<0\),
\(∴α\)为第二象限角,\(∴\cos α<0\),
\(∴\cos α=-\dfrac{\sqrt{5}}{5}\).
方法2 \(∵\tan α=-2\),构造直角三角形\(Rt△ABC\)如下图,
在直角三角形中, \(\cos \alpha=\dfrac{B C}{A C}=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\),
\(∵α∈(0,π)\),且\(\tan α=-2<0\),
\(∴α\)为第二象限角,
\(∴\cos α<0\), \(∴\cos α=-\dfrac{\sqrt{5}}{5}\).
点拨
① 若知\(\sin α\)、\(\cos α\)、\(\tan α\)三者中一个的值,可求另外两个的值,即“知一得二”;
② 在非解答题中用方法二解题速度更快些,只是要多留意三角函数的符号.
【典题2】 已知\(\tan α=-2\),求下列各式的值:
(1) \(\dfrac{4 \sin \alpha-2 \cos \alpha}{5 \cos \alpha+3 \sin \alpha}\); (2) \(\dfrac{1}{4} \sin ^2 \alpha+\dfrac{2}{5} \cos ^2 \alpha\).
解析 方法1 由\(\tan α=-2\),得\(\sin α=-2\cos α\).
(1) \(\dfrac{4 \sin \alpha-2 \cos \alpha}{5 \cos \alpha+3 \sin \alpha}=\dfrac{-8 \cos \alpha-2 \cos \alpha}{5 \cos \alpha-6 \cos \alpha}=10\).
(2) \(\dfrac{1}{4} \sin ^2 \alpha+\dfrac{2}{5} \cos ^2 \alpha=\dfrac{\dfrac{1}{4} \sin ^2 \alpha+\dfrac{2}{5} \cos ^2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=\dfrac{\cos ^2 \alpha+\dfrac{2}{5} \cos ^2 \alpha}{4 \cos ^2 \alpha+\cos ^2 \alpha}=\dfrac{7}{25}\).
方法2 \(∵\tan α=-2\),\(∴\cos α≠0\).
(1) \(\dfrac{4 \sin \alpha-2 \cos \alpha}{5 \cos \alpha+3 \sin \alpha}=\dfrac{4 \tan \alpha-2}{5+3 \tan \alpha}=\dfrac{4 \times(-2)-2}{5+3 \times(-2)}=10\).
(2) \(\dfrac{1}{4} \sin ^2 \alpha+\dfrac{2}{5} \cos ^2 \alpha=\dfrac{\dfrac{1}{4} \sin ^2 \alpha+\dfrac{2}{5} \cos ^2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=\dfrac{\dfrac{1}{\tan ^2 \alpha+\frac{2}{5}}}{\tan ^2 \alpha+1}=\dfrac{7}{25}\).
点拨 方法2中当分式\(\dfrac{f(x)}{g(x)}\)中\(f(x)\)、\(g(x)\)关于\(\cos α\)、\(\sin α\)的“齐次化式”,分子分母同除以\(\cos α\)或\(\sin α\).
第(2)问中巧妙的使用了\(\sin ^2 α+\cos ^2 α=1\).
【典题3】 已知 \(\sin \alpha+\cos \alpha=-\dfrac{1}{5}\).
(1)求\(\sin α\cdot \cos α\)的值;
(2)若\(\dfrac{π}{2}<α<π\),求\(\dfrac{1}{\sin \alpha}-\dfrac{1}{\cos (2 \pi+\alpha)}\)的值.
解析 (1)\(∵\sin \alpha+\cos \alpha=-\dfrac{1}{5}\) ①, \(\therefore(\sin \alpha+\cos \alpha)^2=\dfrac{1}{25}\),
即\(1+2 \sin \alpha \cos \alpha=\dfrac{1}{25}\), \(\therefore \sin \alpha \cdot \cos \alpha=-\dfrac{12}{25}\) ,
(2)由(1)得,\((\sin \alpha-\cos \alpha)^2=1-2 \sin \alpha \cos \alpha=\dfrac{49}{25}\),
又 \(\dfrac{π}{2}<α<π\),\(∴\sin α-\cos α>0\),
\(\therefore \sin \alpha-\cos \alpha=\dfrac{7}{5}\) ②.
\(\therefore \dfrac{1}{\sin \alpha}-\dfrac{1}{\cos (2 \pi+\alpha)}=\dfrac{1}{\sin \alpha}-\dfrac{1}{\cos \alpha}=\dfrac{\cos \alpha-\sin \alpha}{\sin \alpha \cos \alpha}=\dfrac{35}{12}\) .
点拨
① \((\sin α+\cos α)^2=1+2 \sin α \cdot \cos α\); \((\sin α-\cos α)^2=1-2 \sin α \cdot \cos α\).
②\(\sin θ+\cos θ\)、\(\sin θ-\cos θ\)、\(\sin θ\cos θ\)也是“知一得二”.
【巩固练习】
1.已知\(α\)是第四象限角, \(\cos \alpha=\dfrac{12}{13}\),则\(\sin α\)等于( )
A. \(\dfrac{5}{13}\) \(\qquad \qquad \qquad \qquad\) B. \(-\dfrac{5}{13}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{5}{12}\) \(\qquad \qquad \qquad \qquad\) D. \(-\dfrac{5}{12}\)
2.已知\(\alpha=-\dfrac{1}{2}\),则\(\dfrac{2 \sin \alpha \cos \alpha}{\sin ^2 \alpha-\cos ^2 \alpha}\)的值是( )
A. \(\dfrac{4}{3}\) \(\qquad \qquad \qquad \qquad\) B.\(3\) \(\qquad \qquad \qquad \qquad\) C. \(-\dfrac{4}{3}\) \(\qquad \qquad \qquad \qquad\) D.\(-3\)
3.如果角\(θ\)满足\(\sin \theta+\cos \theta=\sqrt{2}\),那么\(\tan \theta+\dfrac{1}{\tan \theta}\)的值是( )
A.\(-1\) \(\qquad \qquad \qquad \qquad\) B.\(-2\) \(\qquad \qquad \qquad \qquad\) C.\(1\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
4.若\(\alpha \in\left(\dfrac{\pi}{2}, \pi\right)\),且\(\cos ^2 \alpha-\sin \alpha=\dfrac{1}{4}\),则\(\tan α\)的值等于( )
A. \(-\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) C. \(\sqrt{3}\) \(\qquad \qquad \qquad \qquad\) D. \(-\sqrt{3}\)
5.已知\(\tan \alpha=\dfrac{4}{3}\),且\(α\)是第三象限角,求\(\sin α\),\(\cos α\)的值.
6.已知\(\sin θ\)、\(\cos θ\)是关于\(x\)的方程 \(x^2-2 \sqrt{2} a x+a=0\)的两个根.
(1)求实数\(a\)的值;(2)若\(θ∈\left(-\dfrac{\pi}{2},0\right)\),求\(\sin θ-\cos θ\)的值.
参考答案
-
答案 \(B\)
解析 \(∵α\)是第四象限角,
\(\therefore \sin \alpha=-\sqrt{1-\cos ^2 \alpha}=-\sqrt{1-\left(\dfrac{12}{13}\right)^2}=-\dfrac{5}{13}\). -
答案 \(A\)
解析 原式 \(=\dfrac{2 \tan \alpha}{\tan ^2 \alpha-1}=\dfrac{2 \times\left(-\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^2-1}=\dfrac{4}{3}\). -
答案 \(D\)
解析 \(\because \sin \theta+\cos \theta=\sqrt{2}\),
\(\therefore 1+2 \sin \theta \cos \theta=2\),即 \(\sin \theta \cos \theta=\dfrac{1}{2}\),
那么 \(\tan \theta+\dfrac{1}{\tan \theta}=\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}=\dfrac{1}{\sin \theta \cos \theta}=2\),
故选:\(D\). -
答案 \(A\)
解析 \(∵\cos ^2 \alpha-\sin \alpha=\dfrac{1}{4}\),\(∴4(1-\sin ^2α )-4\sin α-1=0\),
即\(A\)\(4 \sin ^2α+4\sin α-3=0\),\(∴\)解得 \(\sin \alpha=\dfrac{1}{2}\)或 \(\sin \alpha=-\dfrac{3}{2}\)(舍).
\(∵\alpha \in\left(\dfrac{\pi}{2}, \pi\right)\), \(\therefore \alpha=\dfrac{5 \pi}{6}\), \(\therefore \tan \alpha=\tan \dfrac{5 \pi}{6}=-\dfrac{\sqrt{3}}{3}\).
故选:\(A\). -
答案 \(\cos \alpha=-\dfrac{3}{5}\), \(\sin \alpha=-\dfrac{4}{5}\).
解析 由 \(\tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{4}{3}\)得\(\sin α=\dfrac{4}{3} \cos α\).①
又\(\sin ^2 α+\cos ^2 α=1\),②
由①②得 \(\dfrac{16}{9} \cos ^2 \alpha+\cos ^2 \alpha=1\),即 \(\cos ^2 \alpha=\dfrac{9}{25}\).
\(∵α\)在第三象限,
\(\therefore \cos \alpha=-\dfrac{3}{5}\), \(\sin \alpha=\dfrac{4}{3} \cos \alpha=-\dfrac{4}{5}\). -
答案 (1) \(a=-\dfrac{1}{4}\)或 \(\dfrac{1}{2}\);(2) \(-\dfrac{\sqrt{6}}{2}\).
解析 (1)\(∵\sin θ\)、\(\cos θ\)是关于\(x\)的方程 \(x^2-2 \sqrt{2} a x+a=0\)的两个根,
\(\therefore \sin \theta+\cos \theta=2 \sqrt{2} a\) ①,\(\sin θ\cos θ=a\) ②,
\(△=b^2-4ac=8a^2-4a≥0\),即\(a≤0\)或\(a≥\dfrac{1}{2}\),
\(∴(\sin θ+\cos θ)^2=1+2\sin θ\cos θ=1+2a=8a^2\),
即\(8a^2-2a-1=0\),解得\(a=-\dfrac{1}{4}\)或 \(\dfrac{1}{2}\).
(2)\(∵θ∈\left(-\dfrac{\pi}{2},0 \right)\),
\(∴\sin θ<0\),\(\cos θ>0\),可得\(\sin θ\cos θ=a<0\),由(1)可得\(a=-\dfrac{1}{4}\),
\(∴\sin θ\cos θ=-\dfrac{1}{4}\),
\(∴(\sin θ-\cos θ)^2=1-2\sin θ\cos θ=1+\dfrac{1}{2}=\dfrac{3}{2}\),
又\(\sin θ-\cos θ<0\),\(∴ \sin θ-\cos θ=-\dfrac{\sqrt{6}}{2}\).
(注意判断\(\sin θ-\cos θ\)的正负)
【题型2】三角函数式的化简
【典题1】 化简下列各式:
(1) \(\dfrac{\sqrt{1-2 \sin 130^{\circ} \cos 130^{\circ}}}{\sin 130^{\circ}+\sqrt{1-\sin ^2 130^{\circ}}}\);
(2) \(\sin ^2 \alpha \cdot \tan \alpha+2 \sin \alpha \cdot \cos \alpha+\dfrac{\cos ^2 \alpha}{\tan \alpha}\).
解析 (1)原式 \(=\dfrac{\sqrt{\sin ^2 130^{\circ}-2 \sin 130^{\circ} \cos 130^{\circ}+\cos ^2 130^{\circ}}}{\sin 130^{\circ}+\sqrt{\cos ^2 130^{\circ}}}\)
\(=\dfrac{\left|\sin 130^{\circ}-\cos 130^{\circ}\right|}{\sin 130^{\circ}+\left|\cos 130^{\circ}\right|}=\dfrac{\sin 130^{\circ}-\cos 130^{\circ}}{\sin 130^{\circ}-\cos 130^{\circ}}=1\).
(2)原式 \(=\sin ^2 \alpha \cdot \dfrac{\sin \alpha}{\cos \alpha}+2 \sin \alpha \cos \alpha+\cos ^2 \alpha \cdot \dfrac{\cos \alpha}{\sin \alpha}\)
\(=\dfrac{\sin ^4 \alpha+2 \sin ^2 \alpha \cos ^2 \alpha+\cos ^4 \alpha}{\cos \alpha \sin \alpha}\)\(=\dfrac{\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2}{\sin \alpha \cos \alpha}=\dfrac{1}{\sin \alpha \cos \alpha}\).
点拨 遇到正切一般化为正弦与余弦,"切化弦"的方法.
【巩固练习】
1.若角\(α\)的终边在第二象限,则\(\dfrac{\sin \alpha}{\sqrt{1-\sin ^2 \alpha}}+\dfrac{\sqrt{1-\cos ^2 \alpha}}{\cos \alpha}\) 的值等于( )
A.\(2\) \(\qquad \qquad \qquad \qquad\) B.\(-2\) \(\qquad \qquad \qquad \qquad\) C.\(0\) \(\qquad \qquad \qquad \qquad\) D.\(-2\)或\(2\)
参考答案
- 答案 \(C\)
解析 \(∵α\)是第二象限角,
\(∴\sin α>0\),\(\cos α<0\).
\(\therefore \dfrac{\sin \alpha}{\sqrt{1-\sin ^2 \alpha}}+\dfrac{\sqrt{1-\cos ^2 \alpha}}{\cos \alpha}=\dfrac{\sin \alpha}{|\cos \alpha|}+\dfrac{|\sin \alpha|}{\cos \alpha}=-\tan \alpha+\tan \alpha=0\).
【题型3】三角恒等式的证明
【典题1】 求证: \(\dfrac{1-2 \sin 2 x \cos 2 x}{\cos ^2 2 x-\sin ^2 2 x}=\dfrac{1-\tan 2 x}{1+\tan 2 x}\).
证明 左边 \(=\dfrac{\cos ^2 2 x+\sin ^2 2 x-2 \sin 2 x \cos 2 x}{\cos ^2 2 x-\sin ^2 2 x}=\dfrac{(\cos 2 x-\sin 2 x)^2}{(\cos 2 x-\sin 2 x)(\cos 2 x+\sin 2 x)}\)
\(=\dfrac{\cos 2 x-\sin 2 x}{\cos 2 x+\sin 2 x}=\dfrac{1-\tan 2 x}{1+\tan 2 x}=\)右边,
所以原式成立.
【巩固练习】
1.求证: \(2(1-\sin \alpha)(1+\cos \alpha)=(1-\sin \alpha+\cos \alpha)^2\).
参考答案
- 证明 右边\(=[(1-\sin α)+\cos α]^2\)
\(=(1-\sin α)^2 +\cos ^2α+2 \cos α(1-\sin α )\)
\(=1-2\sin α+\sin ^2 α+\cos ^2 α+2\cos α(1-\sin α)\)
\(=2-2\sin α+2\cos α(1-\sin α)\)
\(=2(1-\sin α)(1+\cos α)=\)左边,
所以原式成立.
分层练习
【A组---基础题】
1.已知\(\cos α=-\dfrac{12}{13}\),\(α∈(π,2π)\),则\(\tan α=\) ( )
A. \(\dfrac{12}{5}\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{5}{12}\) \(\qquad \qquad \qquad \qquad\) C. \(-\dfrac{5}{12}\) \(\qquad \qquad \qquad \qquad\) D. \(-\dfrac{12}{5}\)
2.已知\(\tan \alpha=\dfrac{3}{4}\),则\(\dfrac{\sin \alpha-2 \cos \alpha}{2 \sin \alpha+\cos \alpha}=\)( )
A.\(-2\) \(\qquad \qquad \qquad \qquad\) B.\(2\) \(\qquad \qquad \qquad \qquad\) C.\(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) D. \(\dfrac{1}{2}\)
3.如果角\(θ\)满足\(\sin θ+\cos θ=\sqrt{2}\),那么\(\tan \theta+\dfrac{1}{\tan \theta}\)的值是( )
A.\(-1\) \(\qquad \qquad \qquad \qquad\) B.\(-2\) \(\qquad \qquad \qquad \qquad\) C.\(1\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
4.已知角\(α\)是锐角,若\(\sin α\),\(\cos α\)是关于\(x\)的方程\(x^2+mx+n=0\)的两个实数根,则实数\(m\)和\(n\)一定满足( )
A.\(m^2-4n=0\) \(\qquad \qquad\) B.\(m^2=2n+1\) \(\qquad \qquad\) C.\(m+n+1≤0\) \(\qquad \qquad\) D.\(mn>0\)
5.已知\(α\)是第三象限角,\(4\sin ^2 α-3\sin α\cos α-5\cos ^2 α=1\),则\(\tan α=\)( )
A.\(-1\)或\(2\) \(\qquad \qquad \qquad \qquad\) B. \(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C.\(1\) \(\qquad \qquad \qquad \qquad\) D.\(2\)
6.已知\(α∈(π,\dfrac{3\pi}{2})\),\(\tan α=2\),则\(\cos α=\) \(\underline{\quad \quad}\).
7.已知\(\sin α+\cos α=\dfrac{1}{5}\),则\(\sin α\cos α=\) \(\underline{\quad \quad}\).
8.若\(\sin α=\dfrac{\sqrt{5}}{5}\),则\(\sin ^4 α-\cos ^4 α=\) \(\underline{\quad \quad}\).
9.已知\(α\)为第二象限角,则\(\dfrac{2 \sin \alpha}{\sqrt{1-\cos ^2 \alpha}}+\cos \alpha \sqrt{1+\tan ^2 \alpha}\)的值是\(\underline{\quad \quad}\).
10.已知\(\tan \theta+\dfrac{1}{\tan \theta}=3\),求\(\tan ^2 \theta+(\sin \theta-\cos \theta)^2+\dfrac{1}{\tan ^2 \theta}\)的值.
参考答案
-
答案 \(B\)
解析 \(∵\cos α=-\dfrac{12}{13}\),\(α∈(π,2π)\),则\(α∈(π,\dfrac{3\pi}{2})\),\(∴\sin α<0\).
又\(\sin ^2 α+\cos ^2 α=1\),
\(\therefore \sin ^2 \alpha=1-\cos ^2 \alpha=\dfrac{25}{169}\),
\(\therefore \sin \alpha=-\dfrac{5}{13}\), \(\therefore \tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{5}{12}\). -
答案 \(C\)
解析 \(∵\tan \alpha=\dfrac{3}{4}\) , \(\therefore \dfrac{\sin \alpha-2 \cos \alpha}{2 \sin \alpha+\cos \alpha}=\dfrac{\tan \alpha-2}{2 \tan \alpha+1}=-\dfrac{1}{2}\).故选:\(C\). -
答案 \(D\)
解析 \(∵\sin θ+\cos θ=\sqrt{2}\),\(∴1+2\sin θ\cos θ=2\),即\(\sin θ\cos θ= \dfrac{1}{2}\),
那么 \(\tan \theta+\dfrac{1}{\tan \theta}=\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}=\dfrac{1}{\sin \theta \cos \theta}=2\),
故选:\(D\). -
答案 \(B\)
解析 由题可得\(\sin α+\cos α=-m\),\(\sin α\cos α=n\),
因为\(1=\sin ^2 α+\cos ^2 α=(\sin α+\cos α)^2-2\sin α\cos α=m^2-2n\),
\(∴m^2=2n+1\),
故选:\(B\). -
答案 \(D\)
解析 由\(4\sin ^2 α-3\sin α\cos α-5\cos ^2 α=1\)
可得 \(\dfrac{4 \sin ^2 \alpha-3 \sin \alpha \cos \alpha-5 \cos ^2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=1\).
分子,分母同时除以\(\cos ^2α\),得 \(\dfrac{4 \tan ^2 \alpha-3 \tan \alpha-5}{\tan ^2 \alpha+1}=1\),
解得\(\tan α=-1\)或\(\tan α=2\).
又\(∵α\)是第三象限角,
\(∴\tan α>0\).\(∴\tan α=2\). -
答案 \(-\dfrac{\sqrt{5}}{5}\)
解析 \(∵α∈\left(π,\dfrac{3\pi}{2}\right)\),\(\tan α=2\),
\(∴\cos α<0\), \(\dfrac{\sin \alpha}{\cos \alpha}=2\).
又\(\sin ^2 α+\cos ^2 α=1\),
\(∴5 \cos ^2α=1\),\(∴\cos α=-\dfrac{\sqrt{5}}{5}\). -
答案 \(-\dfrac{12}{25}\)
解析 \(∵\sin α+\cos α=\dfrac{1}{5}\),
\(∴\)两边平方可得:\(\sin ^2 α+\cos ^2 α+2\sin α\cos α=\dfrac{1}{25}\),
\(∴1+2\sin α\cos α=\dfrac{1}{25}\),则\(\sin α\cos α=-\dfrac{12}{25}\). -
答案 \(-\dfrac{3}{5}\)
解析 \(\sin ^4α-\cos ^4α=(\sin ^2α+\cos ^2α )(\sin ^2α-\cos ^2α )\)
\(=\sin ^2α-\cos ^2α=\sin ^2α-(1-\sin ^2α )\)
\(=2 \sin ^2 \alpha-1=2 \times\left(\dfrac{\sqrt{5}}{5}\right)^2-1=-\dfrac{3}{5}\). -
答案 \(1\)
解析 \(∵α\)为第二象限角,\(\cos α<0\),
\(\therefore \dfrac{2 \sin \alpha}{\sqrt{1-\cos ^2 \alpha}}+\cos \alpha \sqrt{1+\tan ^2 \alpha}=\dfrac{2 \sin \alpha}{\sin \alpha}+\cos \alpha \sqrt{\dfrac{1}{\cos ^2 \alpha}}\)\(=2+\cos \alpha\left|\dfrac{1}{\cos \alpha}\right|=2-1=1\).
故答案为:\(1\). -
答案 \(\dfrac{22}{3}\)
解析 由已知得 \(\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}=3\),
\(\therefore \dfrac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}=3\), \(\therefore \sin \theta \cos \theta=\dfrac{1}{3}\).
\(∴\)原式 \(=\left(\tan \theta+\dfrac{1}{\tan \theta}\right)^2-2+(1-2 \sin \theta \cos \theta)=3^2-2+1-\dfrac{2}{3}=\dfrac{22}{3}\).
【B组---提高题】
1.已知 \(\dfrac{1-\cos x+\sin x}{1+\cos x+\sin x}=-2\),则\(\tan x\)的值为\(\underline{\quad \quad}\).
2.已知\(3\sin α+4\cos α=5\),求\(\tan α\).
3.已知\(0<x<π\),\(\sin x+\cos x=\dfrac{1}{4}\).
①求\(\sin x-\cos x\)的值;②求\(\sin ^3x+\cos ^3x\)的值.
参考答案
-
答案 \(\dfrac{4}{3}\)
解析 由条件,得\(1-\cos x+\sin x=-2-2\cos x-2\sin x\),
整理得:\(3\sin x+\cos x=-3\),
即 \(\cos x=-3\sin x-3\) ①,
代入\(\sin ^2 x+\cos ^2 x=1\)中,得\(\sin ^2 x+(-3\sin x-3)^2=1\),
整理得:\(5\sin ^2 x+9\sin x+4=0\),即\((\sin x+1)(5\sin x+4)=0\),
解得\(\sin x=-1\)(舍)或\(\sin x=-\dfrac{4}{5}\),
把 \(\sin x=-\dfrac{4}{5}\),代入①,得\(\cos x=-\dfrac{3}{5}\),所以\(\tan x=\dfrac{4}{3}\). -
答案 \(\dfrac{3}{4}\)
解析 方法1解方程组法
由\(\left\{\begin{array}{c} 3 \sin \alpha+4 \cos \alpha=5 \\ \sin ^2 \alpha+\cos ^2 \alpha=1 \end{array}\right.\)得\(25 \sin ^2α-30\sin α+9=0\),
解得\(\sin α=\dfrac{3}{5}\),
\(∴\cos α=\dfrac{4}{5}\) , \(∴\tan α=\dfrac{3}{4}\).
方法2 “对偶式”法
设\(4\sin α-3\cos α=x\),等式两边平方得\(16 \sin ^2α-24\sin α\cos α+9 \cos ^2α=x^2\) ①
将\(3\sin α+4\cos α=5\)两边平方,得\(9 \sin ^2α+24\sin α\cos α+16 \cos ^2α=25\) ②
由①+②得,\(25=x^2+25\),解得\(x=0\),
\(∴4\sin α-3\cos α=0\), \(∴4\sin α=3\cos α\), \(∴\tan α=\dfrac{3}{4}\).
方法3 “弦化切”法
将\(3\sin α+4\cos α=5\)两边平方,得\(9 \sin ^2α+24\sin α\cos α+16 \cos ^2α=25\)
即\(\dfrac{9 \sin ^2 \alpha+24 \sin \alpha \cos \alpha+16 \cos ^2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=25\),
即 \(\dfrac{9 \tan ^2 \alpha+24 \tan \alpha+16}{\tan ^2 \alpha+1}=25\),解得 \(\tan α=\dfrac{3}{4}\). -
答案 ① \(\dfrac{\sqrt{31}}{4}\); ② \(\dfrac{47}{128}\).
解析 已知\(0<x<π\), \(\sin x+\cos x=\dfrac{1}{4}\).
所以 \((\sin x+\cos x)^2=\dfrac{1}{16}\),解得 \(\sin x \cos x=-\dfrac{15}{32}\),
所以\(\dfrac{\pi}{2}<x<π\).
① \(\sin x-\cos x=|\sin x-\cos x|=\sqrt{(\sin x+\cos x)^2-4 \sin x \cos x}\)
\(=\sqrt{\dfrac{1}{16}-4 \times\left(-\dfrac{15}{32}\right)}=\dfrac{\sqrt{31}}{4}\).
② \(\sin ^3 x+\cos ^3 x=(\sin x+\cos x)\left(\sin ^2 x-\sin x \cos x+\cos ^2 x\right)\)\(=\dfrac{1}{4} \times\left(1+\dfrac{15}{32}\right)=\dfrac{47}{128}\).
【C组---拓展题】
1.已知函数\(y=\cos ^2x+a\sin x-a^2+2a+5\)有最大值\(2\),求实数\(a\)的值.
参考答案
- 答案 \(a=-\dfrac{4}{3}\)或 \(a=\dfrac{3+\sqrt{21}}{2}\).
解析 \(y=-\sin ^2x+a\sin x-a^2+2a+6\),令\(\sin x=t\),\(t∈[-1,1]\),
则\(y=-t^2+at-a^2+2a+6\),对称轴为\(t=\dfrac{a}{2}\),
当\(\dfrac{a}{2}<-1\),即\(a<-2\)时,\([-1,1]\)是函数\(y\)的递减区间, \(y_{\max }=\left.y\right|_{t=-1}=-a^2+a+5=2 \text {, }\),
得 \(a^2-a-3=0\), \(a=\dfrac{1 \pm \sqrt{13}}{2}\)与\(a<-2\)矛盾;
当\(\dfrac{a}{2}>1\),即\(a>2\)时,\([-1,1]\)是函数\(y\)的递增区间, \(y_{\max }=\left.y\right|_{t=-1}=-a^2+a+5=2\),
得\(a^2-3a-3=0\), \(a=\dfrac{3 \pm \sqrt{21}}{2}\),
而\(a>2\),即 \(a=\dfrac{3 + \sqrt{21}}{2}\);
当\(-1≤\dfrac{a}{2}≤1\) ,即\(-2≤a≤2\)时, \(y_{\max }=\left.y\right|_{t=\dfrac{a}{2}}=-\dfrac{3}{4} a^2+2 a+6=2\),
得\(3a^2-8a-16=0\),\(a=4\)或\(a=-\dfrac{4}{3}\),
而\(-2≤a≤2\),即\(a=-\dfrac{4}{3}\);
\(∴a=-\dfrac{4}{3}\)或 \(a=\dfrac{3+\sqrt{21}}{2}\).