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5.3 诱导公式

时间:2022-12-02 21:34:27浏览次数:73  
标签:cos 5.3 frac 公式 诱导 right alpha pi sin

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必修第一册同步巩固,难度2颗星!

基础知识

诱导公式

利用以上6组公式,最好结合图象,利用对称性和全等三角形进行理解消化.
(1) 公式(一) \(\sin ⁡(α+2kπ)=\sin ⁡α\);\(\cos (α+2kπ)=\cos α\) ;\(\tan (α+2kπ)=\tan α\).
由三角函数的定义易得.
(2) 公式(二) \(\sin (π+α)=-\sin α\);\(\cos (π+α)=-\cos α\); \(\tan (π+α)=\tan α\).
证明 如下图,\(α\)的终边与单位圆交于\(P_1 (x ,y)\),则\(π+α\)的终边与单位圆交于\(P_2 (x ,y)\),
显然\(P_2\)与\(P_1\)关于原点对称,则\(P_2 (-x ,-y)\).
由三角函数的定义,可知\(\sin α=y\),\(\cos α=x\), \(\tan \alpha=\frac{y}{x}\);
\(\sin ⁡(π+α)=-y\),\(\cos ⁡(π+α)=-x\),\(\tan ⁡(π+α)=\frac{y}{x}\);
故\(\sin (π+α)=-\sin α\) ; \(\cos (π+α)=-\cos α\) ; \(\tan (π+α)=\tan α\).
image.png
(3) 公式(三) \(\sin (-α)=-\sin α\) ; \(\cos (-α)=\cos α\) ;\(\tan (-α)=-\tan α\).
image.png 若\(P_1 (x ,y)\),则\(P_3 (x ,-y)\).
(4) 公式(四) \(\sin (π-α)=\sin α\) ;\(\cos (π-α)=-\cos α\) ; \(\tan (π-α)=-\tan α\).
image.png若\(P_1 (x ,y)\),则\(P_4 (-x ,y)\).
(5) 公式(五) \(\sin (\frac{\pi}{2} -α)=\cos α\) ; \(\cos (\frac{\pi}{2} -α)=\sin α\).
image.png若\(P_1 (x ,y)\),则\(P_5 (y ,x)\).
(6) 公式(六) \(\sin (\frac{\pi}{2} +α)=\cos α\) ; \(\cos (\frac{\pi}{2} +α)=-\sin α\).
image.png若\(P_1 (x ,y)\),则\(P_6 (-y ,x)\).
 

诱导公式口诀:奇变偶不变,符号看象限

(奇偶指的是\(\frac{\pi}{2} \cdot n+α\)中整数\(n\)是奇数还是偶数,看象限时把\(α\)看作锐角)
\(\sin \left(\frac{\pi}{2} \cdot n+\alpha\right)=\left\{\begin{array}{l} (-1)^{\frac{\pi}{2}} \sin \alpha, n \text { 为偶数 } \\ (-1)^{\frac{n+1}{2}} \cos \alpha, n \text { 为奇数 } \end{array}\right.\),
\(\cos \left(\frac{\pi}{2} \cdot n+\alpha\right)=\left\{\begin{array}{l} (-1)^{\frac{n}{2}} \cos \alpha, n \text { 为偶数 } \\ (-1)^{\frac{n+1}{2}} \sin \alpha, n \text { 为奇数 } \end{array}\right.\).
 

【例】 利用诱导公式化简以下式子:(1) \(\sin (x+π)\); (2) \(\cos (\frac{\pi}{2} -x)\),(3) \(\sin (-x)\).
解析 (1)
image.png
(2)
image.png
(3)
image.png
 

常见结论

(1) \(A+B=π⇒\sin ⁡A=\sin ⁡B\),\(\cos ⁡A=-\cos ⁡B\);
(2) \(A+B=\frac{\pi}{2} ⇒\sin ⁡A=\cos ⁡B\).
【例】 求\(\cos ⁡150°\)、 \(\sin ⁡120°\)的值.
解 \(\cos ⁡150°=-\cos ⁡30°=-\frac{\sqrt{3}}{2}\),\(\sin ⁡120°=\sin ⁡60°=\frac{\sqrt{3}}{2}\).
 

基本方法

【题型1】求值与化简

【典题1】 求下列各三角函数值:
  (1)\(\sin ⁡(-945^∘ )\);(2)\(\cos (-\frac{16\pi}{3} )\).
解析 (1) 方法1
\(\sin \left(-945^{\circ}\right)=-\sin 945^{\circ}=-\sin \left(225^{\circ}+2 \times 360^{\circ}\right)=-\sin 225^{\circ}\)
\(=-\sin ⁡(180^∘+45^∘ )=\sin ⁡45^∘=\frac{\sqrt{2}}{2}\).
方法2 \(\sin ⁡(-945^∘ )=\sin ⁡(135^∘-3×360^∘ )=\sin ⁡135^∘\)\(=\sin ⁡(180^∘-45^∘ )=\sin ⁡45^∘=\frac{\sqrt{2}}{2}\).
(2) 方法1 \(\cos \left(-\frac{16 \pi}{3}\right)=\cos \frac{16 \pi}{3}=\cos \left(\frac{4 \pi}{3}+4 \pi\right)=\cos \frac{4 \pi}{3}=\cos \left(\pi+\frac{\pi}{3}\right)=-\cos \frac{\pi}{3}=-\frac{1}{2}\).
方法2 \(\cos \left(-\frac{16 \pi}{3}\right)=\cos \left(\frac{2 \pi}{3}-6 \pi\right)=\cos \frac{2 \pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=-\cos \frac{\pi}{3}=-\frac{1}{2}\).
点拨 角度负角化正角,大角化小角,小角化锐角.
 

【典题2】化简: \(\frac{\cos (\theta+4 \pi) \cdot \cos ^2(\theta+\pi) \cdot \sin ^2(\theta+3 \pi)}{\sin (\theta-4 \pi) \cdot \sin (5 \pi+\theta) \cdot \cos ^2(-\pi+\theta)}\).
解析 原式 \(=\frac{\cos \theta \cdot \cos ^2 \theta \cdot \sin ^2 \theta}{\sin \theta \cdot \sin (\pi+\theta) \cdot \cos ^2 \theta}=\frac{\cos ^3 \theta \cdot \sin ^2 \theta}{\sin \theta \cdot(-\sin \theta) \cdot \cos ^2 \theta}\)\(=\frac{\cos \theta \cdot \sin ^2 \theta}{-\sin ^2 \theta}=-\cos \theta\).
 

【巩固练习】

  1. 若\(f(\cos x)=\cos 2x\),则\(f(\sin 15°)=\)(  )
     A.\(\frac{1}{2}\) \(\qquad \qquad\) B.\(\frac{\sqrt{3}}{2}\) \(\qquad \qquad\) C.\(-\frac{1}{2}\) \(\qquad \qquad\) D.\(-\frac{\sqrt{3}}{2}\)
     

2.已知函数 \(f(x)=\cos \frac{x}{2}\),则下列四个等式中成立的个数是\(\underline{\quad \quad}\).
 ①\(f(2π-x)=f(x)\);②\(f(2π+x)=f(x)\);③\(f(-x)=-f(x)\);④\(f(-x)=f(x)\).
 

3.化简 \(\frac{\tan (\pi-\alpha) \cos \left(-\frac{\pi}{2}-\alpha\right) \cos (6 \pi-\alpha)}{\sin \left(\frac{\pi}{2}-\alpha\right) \cos \left(\frac{3 \pi}{2}-\alpha\right)}=\) \(\underline{\quad \quad}\).
 

4.化简 \(\frac{\cos \left(90^{\circ}-\alpha\right)}{\sin \left(270^{\circ}+\alpha\right)} \cdot \sin \left(180^{\circ}-\alpha\right) \cdot \cos \left(360^{\circ}-\alpha\right)=\) \(\underline{\quad \quad}\).
 

5.化简: \(\sin \left(2 n \pi+\frac{2 \pi}{3}\right) \cos \left(n \pi+\frac{4 \pi}{3}\right)(n \in \mathrm{Z})\).
 
 

参考答案

  1. 答案 \(D\)
    解析 \(f(\sin ⁡15^∘ )=f(\cos ⁡75^∘ )=\cos 150^∘=\cos (180^∘-30^∘ )\)\(=-\cos 30^∘=-\frac{\sqrt{3}}{2}\).

  2. 答案 \(1\)
    解析 \(f(2 \pi-x)=\cos \frac{2 \pi-x}{2}=\cos \left(\pi-\frac{x}{2}\right)=-\cos \frac{x}{2}=-f(x)\),①不成立;
    \(f(2 \pi+x)=\cos \frac{2 \pi+x}{2}=\cos \left(\pi+\frac{x}{2}\right)=-\cos \frac{x}{2}=-f(x)\),②不成立;
    \(f(-x)=\cos \left(-\frac{x}{2}\right)=\cos \frac{x}{2}=f(x)\),③不成立;④成立.

  3. 答案 \(-\tan α\)
    解析 原式 \(=\frac{-\tan \alpha \cos \left(\frac{\pi}{2}+\alpha\right) \cos (-\alpha)}{\cos \alpha \cos \left[\pi+\left(\frac{\pi}{2}-\alpha\right)\right]}=\frac{\tan \alpha \cdot \sin \alpha \cdot \cos \alpha}{-\cos \alpha \cdot \cos \left(\frac{\pi}{2}-\alpha\right)}=-\frac{\tan \alpha \cdot \sin \alpha}{\sin \alpha}=-\tan \alpha\).

  4. 答案 \(-\sin ^2 α\)
    解析 原式 \(=\frac{\sin \alpha}{-\cos \alpha} \cdot \sin \alpha \cdot \cos \alpha=-\sin ^2 \alpha\).

  5. 答案 \(\frac{\sqrt{3}}{4}\)或 \(-\frac{\sqrt{3}}{4}\)
    解析 (1)当\(n\)为奇数时,
    原式 \(=\sin \frac{2}{3} \pi \cdot\left(-\cos \frac{4}{3} \pi\right)=\sin \left(\pi-\frac{\pi}{3}\right) \cdot\left[-\cos \left(\pi+\frac{\pi}{3}\right)\right]\)\(=\sin \frac{\pi}{3} \cdot \cos \frac{\pi}{3}=\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}}{4}\).
    (2)当n为偶数时,
    原式 \(=\sin \frac{2}{3} \pi \cdot \cos \frac{4}{3} \pi=\sin \left(\pi-\frac{\pi}{3}\right) \cdot \cos \left(\pi+\frac{\pi}{3}\right)\)\(=\sin \frac{\pi}{3} \cdot\left(-\cos \frac{\pi}{3}\right)=\frac{\sqrt{3}}{2} \times\left(-\frac{1}{2}\right)=-\frac{\sqrt{3}}{4}\).
     

【题型2】诱导公式的应用

【典题1】 已知\(\cos ⁡(α-75^∘ )=-\frac{1}{3}\),且\(α\)为第四象限角,求\(\sin (105°+α)\)的值.
解析 \(∵\cos ⁡(α-75^∘ )=-\frac{1}{3}<0\),且\(α\)为第四象限角,
\(∴α-75°\)是第三象限角.
\(\therefore \sin \left(\alpha-75^{\circ}\right)=-\sqrt{1-\cos ^2\left(\alpha-75^{\circ}\right)}=-\sqrt{1-\left(-\frac{1}{3}\right)^2}=-\frac{2 \sqrt{2}}{3}\).
\(\therefore \sin \left(105^{\circ}+\alpha\right)=\sin \left[180^{\circ}+\left(\alpha-75^{\circ}\right)\right]=-\sin \left(\alpha-75^{\circ}\right)=\frac{2 \sqrt{2}}{3}\).
点拨
1.注意已知角\(α-75^∘\)与所求角\(105°+α\)之间的关系,比如它们的和或差、倍数的和差是否为特殊值( \(\frac{\pi}{3}\),\(π\)等).
2.由诱导公式可知:\(A+B=π⇒\sin ⁡A=\sin ⁡B\),\(\cos ⁡A=-\cos ⁡B\);\(A+B=\frac{\pi}{2} ⇒\sin ⁡A=\cos ⁡B\).
 

【典题2】已知\(α∈(\frac{\pi}{2} ,π)\),且\(\sin (π-α)+\cos (2π+α)=\frac{\sqrt{2}}{3}\).求值:
  (1)\(\sin α-\cos α\). (2)\(\tan α\).
解析 已知\(α∈(\frac{\pi}{2} ,π)\),且\(\sin (π-α)+\cos (2π+α)=\frac{\sqrt{2}}{3}\),即\(\sin α+\cos α=\frac{\sqrt{2}}{3}\),
平方可得\(1+2\sin α\cos α=\frac{2}{9}\),即\(2\sin α\cos α=-\frac{7}{9}\).
\(\therefore \sin \alpha-\cos \alpha=\sqrt{(\sin \alpha-\cos \alpha)^2}=\sqrt{1-2 \sin \alpha \cos \alpha}=\frac{4}{3}\).
(2) \(\because \sin \alpha+\cos \alpha=\frac{\sqrt{2}}{3}\), \(\sin \alpha-\cos \alpha=\frac{4}{3}\), \(\therefore \sin \alpha=\frac{4+\sqrt{2}}{6}\), \(\cos \alpha=\frac{\sqrt{2}-4}{6}\),
\(\therefore \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{4+\sqrt{2}}{\sqrt{2}-4}=-\frac{9+4 \sqrt{2}}{7}\).
 

【典题3】证明 \(\frac{\sin \left(\alpha+\frac{3 \pi}{2}\right) \sin \left(\frac{3 \pi}{2}-\alpha\right) \cdot \tan ^2(-\alpha) \cdot \tan (\pi-\alpha)}{\cos \left(\frac{\pi}{2}-\alpha\right) \cos \left(\frac{\pi}{2}+\alpha\right)}=\tan \alpha\).
证明 左边 \(=\frac{-\cos \alpha \cdot(-\cos \alpha) \tan ^2 \alpha \cdot(-\tan \alpha)}{\sin \alpha \cdot(-\sin \alpha)}=\frac{\tan ^2 \alpha \tan \alpha}{\tan ^2 \alpha}=\tan \alpha=\)右边.
\(∴\)原式成立.
 

【巩固练习】

1.已知\(α∈(0,π)\),且\(\cos ⁡(π+α)=-\frac{\sqrt{3}}{2}\),则\(\sin ⁡α=\) ( )
  A. \(\frac{1}{2}\) \(\qquad \qquad\) B. \(-\frac{1}{2}\) \(\qquad \qquad\) C. \(\frac{\sqrt{3}}{2}\) \(\qquad \qquad\) D. \(-\frac{\sqrt{3}}{2}\)
 

2.若 \(\cos \left(\frac{\pi}{6}+\alpha\right)=-\frac{1}{3}\),那么 \(\sin \left(\frac{2 \pi}{3}+\alpha\right)\)的值为(  )
 A.\(-\frac{1}{3}\) \(\qquad \qquad\) B.\(\frac{1}{3}\) \(\qquad \qquad\) C.\(-\frac{2 \sqrt{2}}{3}\) \(\qquad \qquad\) D.\(\frac{2 \sqrt{2}}{3}\)
 

3.已知 \(\sin \left(\frac{\pi}{3}-\alpha\right)=\frac{1}{2}\),求 \(\cos \left(\frac{\pi}{6}+\alpha\right)=\) \(\underline{\quad \quad}\).
 

4.证明: \(\frac{\sin \left(\frac{\pi}{2}-\alpha\right) \cos \left(\frac{\pi}{2}+\alpha\right)}{\cos (\pi+\alpha)}-\frac{\sin (2 \pi-\alpha) \cos \left(\frac{\pi}{2}-\alpha\right)}{\sin (\pi-\alpha)}=2 \sin \alpha\).
 

5.已知函数 \(f(\alpha)=\frac{\sin \left(\alpha-\frac{\pi}{2}\right) \cos \left(\frac{3 \pi}{2}+\alpha\right) \tan (2 \pi-\alpha)}{\tan (\alpha+\pi) \sin (\alpha+\pi)}\).
(1)化简\(f(α)\);
(2)若 \(f(\alpha) \cdot f\left(\alpha+\frac{\pi}{2}\right)=-\frac{1}{8}\),且 \(\frac{\pi}{4} \leq \alpha \leq \frac{\pi}{2}\),求 \(f(\alpha)+f\left(\alpha+\frac{\pi}{2}\right)\);
(3)若 \(f\left(\alpha+\frac{\pi}{2}\right)=2 f(\alpha)\),求 \(f(\alpha) \cdot f\left(\alpha+\frac{\pi}{2}\right)\).
 
 

参考答案

  1. 答案 \(A\)
    解析 因为\(α∈(0,π)\),且\(\cos ⁡(π+α)=-\cos ⁡α=-\frac{\sqrt{3}}{2}\),可得\(\cos ⁡α=\frac{\sqrt{3}}{2}\),
    所以可得\(\sin ⁡α=\sqrt{1-\cos ^2⁡α}=\frac{1}{2}\),故选\(A\).

  2. 答案 \(A\)
    解析 \(\sin \left(\frac{2 \pi}{3}+\alpha\right)=\sin \left[\frac{\pi}{2}+\left(\frac{\pi}{6}+\alpha\right)\right]=\cos \left(\frac{\pi}{6}+\alpha\right)=-\frac{1}{3}\).

  3. 答案 \(\frac{1}{2}\)
    解析 \(\cos \left(\frac{\pi}{6}+\alpha\right)=\cos \left[\frac{\pi}{2}-\left(\frac{\pi}{3}-\alpha\right)\right]=\sin \left(\frac{\pi}{3}-\alpha\right)=\frac{1}{2}\).

  4. 证明 左边 \(=\frac{\cos \alpha(-\sin \alpha)}{-\cos \alpha}-\frac{\sin (-\alpha) \sin \alpha}{\sin \alpha}=\sin \alpha-(-\sin \alpha)=2 \sin \alpha=\)右边,所以原式成立.

  5. 答案 (1) \(-\cos α\)(2) \(-\frac{\sqrt{3}}{2}\)(3) \(\frac{2}{5}\)
    解析 (1) \(f(\alpha)=\frac{(-\cos \alpha) \sin \alpha(-\tan \alpha)}{\tan \alpha(-\sin \alpha)}=-\cos \alpha\);
    (2) \(f\left(\alpha+\frac{\pi}{2}\right)=-\cos \left(\alpha+\frac{\pi}{2}\right)=\sin \alpha\),
    因为 \(f(\alpha) \cdot f\left(\alpha+\frac{\pi}{2}\right)=-\frac{1}{8}\),所以 \(\cos \alpha \cdot \sin \alpha=\frac{1}{8}\),
    可得 \((\sin \alpha-\cos \alpha)^2=\frac{3}{4}\),结合 \(\frac{\pi}{4} \leq \alpha \leq \frac{\pi}{2}\),\(\cos α>\sin α\),
    所以 \(f(\alpha)+f\left(\alpha+\frac{\pi}{2}\right)=\sin \alpha-\cos \alpha=-\frac{\sqrt{3}}{2}\),
    (3) 若\(f(α+\frac{\pi}{2})=2f(α)\),则\(\sin ⁡α=-2 \cos ⁡α\),代入\(\sin ^2⁡α+\cos ^2⁡α=1\),
    解得 \(\cos ^2 \alpha=\frac{1}{5}\),
    所以 \(f(\alpha) \cdot f\left(\alpha+\frac{\pi}{2}\right)=-\sin \alpha \cos \alpha=2 \cos ^2 \alpha=\frac{2}{5}\).
     

分层练习

【A组---基础题】

1.\(\cos 300°=\)(  )
  A.\(-\frac{\sqrt{3}}{2}\) \(\qquad \qquad\) B.\(-\frac{1}{2}\) \(\qquad \qquad\) C.\(\frac{1}{2}\) \(\qquad \qquad\) D.\(\frac{\sqrt{3}}{2}\)
 

2.已知\(f(x)=\sin x\),下列式子成立的是(  )
 A.\(f(x+π)=\sin x\) \(\qquad \qquad\) B.\(f(2π-x)=\sin x\)
\(\qquad \qquad\) C.\(f(x-\frac{\pi}{2})=-\cos x\) \(\qquad \qquad\) D.\(f(π-x)=-f(x)\)
 

3.设\(\tan (5π+α)=m\),则 \(\frac{\sin (\alpha-3 \pi)+\cos (\pi-\alpha)}{\sin (-\alpha)-\cos (\pi+\alpha)}\)的值为(  )
 A. \(\frac{m+1}{m-1}\) \(\qquad \qquad\) B. \(\frac{m-1}{m+1}\) \(\qquad \qquad\) C.\(-1\) \(\qquad \qquad\) D.\(1\)
 

4.若 \(\text { isin }\left(\alpha+\frac{3 \pi}{2}\right)=\frac{3}{5}\),且\(α\)是第三象限角, 则 \(\cos \left(\alpha+\frac{2021 \pi}{2}\right)=\) (  )
  A. \(\frac{3}{5}\) \(\qquad \qquad\) B. \(-\frac{3}{5}\) \(\qquad \qquad\) C. \(\frac{4}{5}\) \(\qquad \qquad\) D. \(-\frac{4}{5}\)
 

5.已知 \(\sin \left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}\),则 \(\cos \left(\frac{\pi}{4}+\alpha\right)\)的值等于(  )
 A.\(\frac{2 \sqrt{2}}{3}\) \(\qquad \qquad\) B.\(-\frac{2 \sqrt{2}}{3}\) \(\qquad \qquad\) C.\(\frac{1}{3}\) \(\qquad \qquad\) D.\(-\frac{1}{3}\)
 

6.已知\(\cos α=\frac{1}{5}\),且\(α\)为第四象限角,那么 \(\cos \left(\alpha+\frac{\pi}{2}\right)=\) \(\underline{\quad \quad}\)\(\underline{\quad \quad}\).
 

7.化简 \(\sin (\pi+\alpha) \cos \left(\frac{3 \pi}{2}+\alpha\right)+\sin \left(\frac{\pi}{2}+\alpha\right) \cos (\pi+\alpha)=\) \(\underline{\quad \quad}\).
 

8.若 \(\sin (3 \pi+\theta)=\frac{1}{4}\),求 \(\frac{\cos (\pi+\theta)}{\cos (-\pi+\theta)[\cos (\pi+\theta)-1]}-\frac{\cos (\theta-2 \pi)}{\cos (\theta+2 \pi) \cos (\theta+\pi)+\cos (-\theta)}\)的值.
 

9.已知 \(\cos \left(\frac{\pi}{6}-\alpha\right)=\frac{\sqrt{3}}{3}\),求 \(\cos \left(\frac{5 \pi}{6}+\alpha\right)-\sin ^2\left(\alpha-\frac{\pi}{6}\right)\)的值.
 

10.已知 \(\sin (\alpha+\pi)=\frac{4}{5}\),且\(\sin α \cdot \cos α<0\),求 \(\frac{2 \sin (\alpha-\pi)+3 \tan (3 \pi-\alpha)}{4 \cos (\alpha-3 \pi)}\)的值.
 
 

参考答案

  1. 答案 \(C\)

  2. 答案 \(C\)

  3. 答案 \(A\)

  4. 答案 \(C\)
    解析 \(\because \sin \left(\alpha+\frac{3 \pi}{2}\right)=-\cos \alpha=\frac{3}{5}\), \(\therefore \cos \alpha=-\frac{3}{5}\),
    又\(α\)是第三象限角, \(\therefore \sin \alpha=-\sqrt{1-\cos ^2 \alpha}=-\frac{4}{5}\),
    \(\therefore \cos \left(\alpha+\frac{2021 \pi}{2}\right)=-\sin \alpha=\frac{4}{5}\).
    故选\(C\).

  5. 答案 \(D\)
    解析 \(\because \frac{\pi}{4}+\alpha-\left(\alpha-\frac{\pi}{4}\right)=\frac{\pi}{2}\),
    \(\therefore \cos \left(\frac{\pi}{4}+\alpha\right)=\cos \left[\frac{\pi}{2}+\left(\alpha-\frac{\pi}{4}\right)\right]=-\sin \left(\alpha-\frac{\pi}{4}\right)=-\frac{1}{3}\).故选\(D\).

  6. 答案 \(\frac{2 \sqrt{6}}{5}\)
    解析 \(∵α\)为第四象限 角, \(\therefore \sin \alpha=-\sqrt{1-\cos ^2 \alpha}=-\frac{2 \sqrt{6}}{5}\),
    从而 \(\cos \left(\alpha+\frac{\pi}{2}\right)=-\sin \alpha=\frac{2 \sqrt{6}}{5}\).

  7. 答案 \(-1\)
    解析 原式\(=-\sin α·\sin α-\cos α·\cos α=-1\).

  8. 答案 \(-32\)
    解析 \(\because \sin (3 \pi+\theta)=\frac{1}{4}\), \(\therefore \sin (\pi+\theta)=\frac{1}{4}\).
    \(\therefore \sin (\theta)=-\frac{1}{4}\).
    \(\therefore \frac{\cos (\pi+\theta)}{\cos (-\pi+\theta)[\cos (\pi+\theta)-1]}-\frac{\cos (\theta-2 \pi)}{\cos (\theta+2 \pi) \cos (\theta+\pi)+\cos (-\theta)}\)
    \(=\frac{\cos \theta}{\cos (\pi-\theta)(1+\cos \theta)}-\frac{\cos \theta}{\cos \theta-\cos ^2 \theta}\)
    \(=\frac{-1}{1+\cos \theta}-\frac{1}{1-\cos \theta}=-\frac{2}{1-\cos ^2 \theta}=-\frac{2}{\sin ^2 \theta}=-32\).

  9. 答案 \(-\frac{2+\sqrt{3}}{3}\)
    解析 \(\cos \left(\frac{5 \pi}{6}+\alpha\right)=\cos \left[\pi-\left(\frac{\pi}{6}-\alpha\right)\right]=-\cos \left(\frac{\pi}{6}-\alpha\right)=-\frac{\sqrt{3}}{3}\),
    而 \(\sin ^2\left(\alpha-\frac{\pi}{6}\right)=1-\cos ^2\left(\frac{\pi}{6}-\alpha\right)=1-\frac{1}{3}=\frac{2}{3}\),
    \(∴\)原式 \(=\frac{-\sqrt{3}}{3}-\frac{2}{3}=-\frac{2+\sqrt{3}}{3}\).

  10. 答案 \(-\frac{7}{3}\)
    解析 \(\because \sin (\alpha+\pi)=\frac{4}{5}\), \(\therefore \sin \alpha=-\frac{4}{5}\).
    又\(\sin α\cos ⁡α<0\),\(∴\cos α>0\), \(\cos \alpha=\sqrt{1-\sin ^2 \alpha}=\frac{3}{5}\),
    \(\therefore \tan \alpha=-\frac{4}{3}\).
    原式 \(=\frac{-2 \sin \alpha-3 \tan \alpha}{-4 \cos \alpha}=\frac{2 \times\left(-\frac{4}{5}\right)+3 \times\left(-\frac{4}{3}\right)}{4 \times \frac{3}{5}}=-\frac{7}{3}\).

 

【B组---提高题】

1.设 \(f(n)=\cos \left(\frac{n \pi}{2}+\frac{\pi}{4}\right)\),则\(f(1)+f(2)+f(3)+⋯+f(2018)\)等于\(\underline{\quad \quad}\).
 

2.已知 \(g(\theta)=\frac{\cos \left(-\theta-\frac{\pi}{2}\right) \cdot \sin \left(\frac{7 \pi}{2}+\theta\right)}{\sin (2 \pi-\theta)}\).
(1)化简\(g(θ)\);
(2)若 \(g\left(\frac{\pi}{3}+\theta\right)=\frac{1}{3}\), \(\theta \in\left(\frac{\pi}{6}, \frac{7 \pi}{6}\right)\),求 \(g\left(\frac{5 \pi}{6}+\theta\right)\)的值;
(3)若 \(g\left(\frac{3}{2} \pi-\theta\right)-g(\theta)=\frac{1}{3}\) , \(\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\),求 \(g(\theta)-g\left(\frac{\pi}{2}-\theta\right)\)的值.
 
 

参考答案

  1. 答案 \(- \sqrt{2}\)
    解析 \(\because f(n+4)=\cos \left[\frac{(n+4) \pi}{2}+\frac{\pi}{4}\right]=\cos \left(\frac{n \pi}{2}+\frac{\pi}{4}\right)\) ,
    \(∴f(n)\)是以\(4\)为周期的函数,
    又 \(f(1)=-\frac{\sqrt{2}}{2}\) , \(f(2)=-\frac{\sqrt{2}}{2}\), \(f(3)=\frac{\sqrt{2}}{2}\), \(f(4)=\frac{\sqrt{2}}{2}\) ,
    \(∴f(1)+f(2)+f(3)+⋯+f(2018)\)
    \(=504[f(1)+f(2)+f(3)+f(4)]+f(1)+f(2)=- \sqrt{2}\).

  2. 答案 (1)\(g(θ)=-\cos θ\); (2) \(\frac{2 \sqrt{2}}{3}\)或\(-\frac{2 \sqrt{2}}{3}\); (3) \(\frac{\sqrt{17}}{3}\)
    解析 (1) \(g(\theta)=\frac{\cos \left(\theta+\frac{\pi}{2}\right) \sin \left(4 \pi-\frac{\pi}{2}+\theta\right)}{\sin (-\theta)}=\frac{-\sin \theta(-\cos \theta)}{-\sin \theta}=-\cos \theta\);
    (2)\(∵ \theta \in\left(\frac{\pi}{6}, \frac{7 \pi}{6}\right)\), \(\therefore \frac{\pi}{3}+\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\),
    \(\because g\left(\frac{\pi}{3}+\theta\right)=-\cos \left(\frac{\pi}{3}+\theta\right)=\frac{1}{3}\) ,即 \(\cos \left(\frac{\pi}{3}+\theta\right)=-\frac{1}{3}\);
    \(\therefore g\left(\frac{5 \pi}{6}+\theta\right)=-\cos \left(\frac{5 \pi}{6}+\theta\right)=-\cos \left(\frac{\pi}{2}+\frac{\pi}{3}+\theta\right)=\sin \left(\frac{\pi}{3}+\theta\right)\);
    \(∴\)当 \(\frac{\pi}{3}+\theta \in\left(\frac{\pi}{2}, \pi\right)\)时,
    \(g\left(\frac{5 \pi}{6}+\theta\right)=\sin \left(\frac{\pi}{3}+\theta\right)=\sqrt{1-\cos ^2\left(\frac{\pi}{3}+\theta\right)}=\frac{2 \sqrt{2}}{3}\);
    当 \(\frac{\pi}{3}+\theta \in\left(\pi, \frac{3 \pi}{2}\right)\),
    \(g\left(\frac{5 \pi}{6}+\theta\right)=\sin \left(\frac{\pi}{3}+\theta\right)=-\sqrt{1-\cos ^2\left(\frac{\pi}{3}+\theta\right)}=-\frac{2 \sqrt{2}}{3}\);
    (3) \(g(\theta)-g\left(\frac{\pi}{2}-\theta\right)=-\cos \theta+\cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta-\cos \theta\)
    由 \(g\left(\frac{3}{2} \pi-\theta\right)-g(\theta)=\frac{1}{3}\),得 \(-\cos \left(\frac{3}{2} \pi-\theta\right)+\cos \theta=\frac{1}{3}\),
    整理得 \(\sin \theta+\cos \theta=\frac{1}{3}\),
    两边平方得: \((\sin \theta+\cos \theta)^2=1+2 \sin \theta \cos \theta=\frac{1}{9}\),即 \(2 \sin \theta \cos \theta=-\frac{8}{9}<0\),
    \(\therefore(\sin \theta-\cos \theta)^2=1-2 \sin \theta \cos \theta=\frac{17}{9} \Rightarrow \sin \theta-\cos \theta=\pm \frac{\sqrt{17}}{3}\),
    \(\because \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\),
    \(∴\cos θ>0\) ,\(\sin θ<0\),即\(\sin θ-\cos θ <0\),
    则 \(g(\theta)-g\left(\frac{\pi}{2}-\theta\right)=\frac{\sqrt{17}}{3}\).
     

【C组---拓展题】

  1. \(\sin ^2 1^{\circ}+\sin ^2 2^{\circ}+\sin ^2 3^{\circ}+\cdots+\sin ^2 89^{\circ}=\)\(\underline{\quad \quad}\).
     
     

参考答案

  1. 答案 \(\frac{89}{2}\)
    解析 设 \(S=\sin ^2 1^{\circ}+\sin ^2 2^{\circ}+\sin ^2 3^{\circ}+\cdots+\sin ^2 89^{\circ}=\) ①
    又 \(\because \mathrm{S}=\sin ^2 89^{\circ}+\sin ^2 88^{\circ}+\sin ^2 87^{\circ}+\cdots+\sin ^2 1^{\circ}\)
    \(=\cos ^2 1^{\circ}+\cos ^2 2^{\circ}+\cos ^2 3^{\circ}+\cdots+\cos ^2 89^{\circ}\) ②
    由①+②得 \(2S=89\),则 \(S=\frac{89}{2}\).

标签:cos,5.3,frac,公式,诱导,right,alpha,pi,sin
From: https://www.cnblogs.com/zhgmaths/p/16945670.html

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