5.3 誘導公式

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模塊導圖

知識剖析

誘導公式

(1) 公式(一) \(\sin⁡(α+2kπ)=\sin⁡α\); \(\cos (α+2kπ)=\cos α\) ; \(\tan (α+2kπ)=\tan α\).
(2) 公式(二) \(\sin (π+α)=-\sin α\) ; \(\cos (π+α)=-\cos α\) ; \(\tan (π+α)=\tan α\).

若\(P_1 (x ,y)\)，則\(P_2 (-x ,-y)\).
(3) 公式(三) \(\sin (-α)=-\sin α\) ; $\cos (-α)=\cos α $; \(\tan (-α)=-\tan α\).

若\(P_1 (x ,y)\)，則\(P_3 (x ,-y)\).
(4) 公式(四) \(\sin (π-α)=\sin α\); \(\cos (π-α)=-\cos α\) ; \(\tan (π-α)=-\tan α\).

若\(P_1 (x ,y)\)，則\(P_4 (-x ,y)\).
(5) 公式(五) \(\sin \left(\frac{\pi}{2}-\alpha\right)=\cos \alpha\)；\(\cos \left(\frac{\pi}{2}-\alpha\right)=\sin \alpha\).

若\(P_1 (x ,y)\)，則\(P_5 (y ,x)\).
(6) 公式(六) \(\sin \left(\frac{\pi}{2}+\alpha\right)=\cos \alpha\) ; \(\cos \left(\frac{\pi}{2}+\alpha\right)=-\sin \alpha\).

若\(P_1 (x ,y)\)，則\(P_6 (-y ,x)\).
利用以上6組公式，最好結合圖象，利用對稱性和全等三角形進行理解消化.

誘導公式口訣：奇變偶不變，符號看象限

(奇偶指的是\(\dfrac{\pi}{2} \cdot n+\alpha\)中整數\(n\)是奇數還是偶數，看象限時把\(α\)看作銳角)
\(\sin \left(\dfrac{\pi}{2} \cdot n+\alpha\right)=\left\{\begin{array}{l} (-1)^{\frac{n}{2}} \sin \alpha, n \text { 為偶數 } \\ (-1)^{\frac{n+1}{2}} \cos \alpha, n \text { 為奇數 } \end{array}\right.\)
\(\cos \left(\dfrac{\pi}{2} \cdot n+\alpha\right)=\left\{\begin{array}{l} (-1)^{\frac{n}{2}} \cos \alpha, n \text { 為偶數 } \\ (-1)^{\frac{n+1}{2}} \sin \alpha, n \text { 為奇數 } \end{array}\right.\)
 

經典例題

【題型一】求具體角度的三角函數值

【典題1】 \(\sin 780^{\circ}+\cos 210^{\circ}+\tan 225^{\circ}\)的值為\(\underline{\quad \quad}\)．
【解析】\(\sin 780^{\circ}+\cos 210^{\circ}+\tan 225^{\circ}\)
\(=\sin \left(720^{\circ}+60^{\circ}\right)+\cos \left(180^{\circ}+30^{\circ}\right)+\tan \left(180^{\circ}+45^{\circ}\right)\)
\(=\sin 60^{\circ}-\cos 30^{\circ}+\tan 45^{\circ}\)
\(=\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}+1=1\).
【點撥】角度負角化正角，大角化小角，小角化銳角.
 

【題型二】誘導公式的運用

【典題1】 設\(f(n)=\cos \left(\dfrac{n \pi}{2}+\dfrac{\pi}{4}\right)\)，
則\(f(1)+f(2)+f(3)+⋯+f(2018)\)等於\(\underline{\quad \quad}\)．
【解析】\(\because f(n+4)=\cos \left[\dfrac{(n+4) \pi}{2}+\dfrac{\pi}{4}\right]\)\(=\cos \left(\dfrac{n \pi}{2}+\dfrac{\pi}{4}\right)\),
\(∴f(n)\)是以\(4\)為周期的函數，
又\(f(1)=-\dfrac{\sqrt{2}}{2}\) ,\(f(2)=-\dfrac{\sqrt{2}}{2}\) ,
\(f(3)=\dfrac{\sqrt{2}}{2}\) ,\(f(4)=\dfrac{\sqrt{2}}{2}\) ,
\(∴f(1)+f(2)+f(3)+⋯+f(2018)\)
\(=504[f(1)+f(2)+f(3)+f(4)]+f(1)+f(2)\)\(=-\sqrt{2}\)．
【點撥】數值比較大項數比較多的時候，注意周期性.
 

【典題2】 已知\(\cos \left(\dfrac{\pi}{6}-\alpha\right)=\dfrac{3}{4}\)，則\(\sin \left(\alpha-\dfrac{2 \pi}{3}\right)=\) \(\underline{\quad \quad}\)．
【解析】\(\because \cos \left(\dfrac{\pi}{6}-\alpha\right)=\dfrac{3}{4}\)，
\(\therefore \sin \left(\alpha-\dfrac{2 \pi}{3}\right)=\sin \left[-\dfrac{\pi}{2}-\left(\dfrac{\pi}{6}-\alpha\right)\right]\)\(=-\cos \left(\dfrac{\pi}{6}-\alpha\right)=-\dfrac{3}{4}\).
【點撥】
① 注意到\(\left(\dfrac{\pi}{6}-\alpha\right)+\left(\alpha-\dfrac{2 \pi}{3}\right)=-\dfrac{\pi}{2}\)是\(\dfrac{\pi}{2}\)的倍數，則可利用誘導公式，這屬於整體代換，相當於令\(\dfrac{\pi}{6}-\alpha=t\).
② 對公式的理解要注意一點:比如誘導公式\(\sin \left(\dfrac{\pi}{2}+\alpha\right)=\cos \alpha\)中的\(α\)其實它可以是一數(如\(\dfrac{\pi}{4}, \dfrac{\pi}{3}\))、一字母(如\(β、θ\))或者一式子(如\(\alpha^{2}\)，\(\beta+\dfrac{\pi}{3}\))，利用公式要特別靈活.
 

【典題3】 已知\(g(\theta)=\dfrac{\cos \left(-\theta-\dfrac{\pi}{2}\right) \cdot \sin \left(\dfrac{7 \pi}{2}+\theta\right)}{\sin (2 \pi-\theta)}\)．
(1)化簡\(g(θ)\)；
(2)若\(g\left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{3}\),\(\theta \in\left(\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\right)\)，求\(g\left(\dfrac{5 \pi}{6}+\theta\right)\)的值；
(3)若\(g\left(\dfrac{3}{2} \pi-\theta\right)-g(\theta)=\dfrac{1}{3}\) ,\(\theta \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)，求\(g(\theta)-g\left(\dfrac{\pi}{2}-\theta\right)\)的值．
【解析】 (1) \(g(\theta)=\dfrac{\cos \left(\theta+\dfrac{\pi}{2}\right) \sin \left(4 \pi-\dfrac{\pi}{2}+\theta\right)}{\sin (-\theta)}\)\(=\dfrac{-\sin \theta(-\cos \theta)}{-\sin \theta}=-\cos \theta\)；
(2)\(\because \theta \in\left(\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\right)\)， \(\therefore \dfrac{\pi}{3}+\theta \in\left(\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right)\)，
\(\because g\left(\dfrac{\pi}{3}+\theta\right)=-\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{3}\)，
即\(\cos \left(\dfrac{\pi}{3}+\theta\right)=-\dfrac{1}{3}\)；
\(\therefore g\left(\dfrac{5 \pi}{6}+\theta\right)=-\cos \left(\dfrac{5 \pi}{6}+\theta\right)\)\(=-\cos \left(\dfrac{\pi}{2}+\dfrac{\pi}{3}+\theta\right)=\sin \left(\dfrac{\pi}{3}+\theta\right)\)；
\(∴\)當\(\dfrac{\pi}{3}+\theta \in\left(\dfrac{\pi}{2}, \pi\right)\)時，
\(g\left(\dfrac{5 \pi}{6}+\theta\right)=\sin \left(\dfrac{\pi}{3}+\theta\right)\)\(=\sqrt{1-\cos ^{2}\left(\dfrac{\pi}{3}+\theta\right)}=\dfrac{2 \sqrt{2}}{3}\)；
當\(\dfrac{\pi}{3}+\theta \in\left(\pi, \dfrac{3 \pi}{2}\right)\)，
\(g\left(\dfrac{5 \pi}{6}+\theta\right)=\sin \left(\dfrac{\pi}{3}+\theta\right)\)\(=-\sqrt{1-\cos ^{2}\left(\dfrac{\pi}{3}+\theta\right)}=-\dfrac{2 \sqrt{2}}{3}\)；
(3) \(g(\theta)-g\left(\dfrac{\pi}{2}-\theta\right)\)\(=-\cos \theta+\cos \left(\dfrac{\pi}{2}-\theta\right)=\sin \theta-\cos \theta\)
由\(g\left(\dfrac{3}{2} \pi-\theta\right)-g(\theta)=\dfrac{1}{3}\)，
得\(-\cos \left(\dfrac{3}{2} \pi-\theta\right)+\cos \theta=\dfrac{1}{3}\)，
整理得\(\sin \theta+\cos \theta=\dfrac{1}{3}\)，
兩邊平方得\((\sin \theta+\cos \theta)^{2}=1+2 \sin \theta \cos \theta=\dfrac{1}{9}\)，
即\(2 \sin \theta \cos \theta=-\dfrac{8}{9}<0\)，
\(\therefore(\sin \theta-\cos \theta)^{2}=1-2 \sin \theta \cos \theta=\dfrac{17}{9}\)
\(\Rightarrow \sin \theta-\cos \theta=\pm \dfrac{\sqrt{17}}{3}\)，
\(\because \theta \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)，
\(\therefore \cos \theta>0, \sin \theta<0\)，

即\(\sin \theta-\cos \theta<0\)，

則\(g(\theta)-g\left(\dfrac{\pi}{2}-\theta\right)=\dfrac{\sqrt{17}}{3}\)．
【點撥】在解題中綜合法與分析法相結合.
 

鞏固練習

1(★) 若\(\sin \alpha=\dfrac{4}{5}\)，則 (　　)
A．\(\cos \left(\dfrac{\pi}{2}-\alpha\right)=\dfrac{4}{5}\) \(\qquad \qquad \qquad\) B．\(\sin \left(\dfrac{\pi}{2}-\alpha\right)=\dfrac{3}{5}\)
C．\(\sin (\pi+\alpha)=\dfrac{4}{5}\) \(\qquad \qquad \qquad\)D．\(\sin (\pi-\alpha)=-\dfrac{4}{5}\)
 

2(★) 在\(△ABC\)中，下列等式一定成立的是 (　　)
A．\(\sin (A+B)=-\sin C\) \(\qquad \qquad \qquad\) B．\(cos(A+B)=cosC\)
C．\(\cos \dfrac{B+C}{2}=\sin \dfrac{A}{2}\) \(\qquad \qquad \qquad\) D．\(\sin \dfrac{B+C}{2}=\sin \dfrac{A}{2}\)
 

3(★)\(\sin \left(-\dfrac{17 \pi}{6}\right)+\cos \left(-\dfrac{20 \pi}{3}\right)+\tan \left(-\dfrac{53 \pi}{6}\right)=\) \(\underline{\quad \quad}\) ．
 

4(★★) 已知\(\sin \left(\alpha-\dfrac{\pi}{3}\right)=\dfrac{1}{3}\)，則\(\cos \left(\dfrac{\pi}{6}+\alpha\right)=\) \(\underline{\quad \quad}\) ．
 

5(★★)已知\(\sinθ,\cosθ\)是關於\(x\)的方程\(x^2-ax+a=0(a∈R)\)的兩個根．
(1)求\(\cos ^{3}\left(\dfrac{\pi}{2}-\theta\right)+\sin ^{3}\left(\dfrac{\pi}{2}-\theta\right)\)的值；
(2)求\(\tan (\pi-\theta)-\dfrac{1}{\tan \theta}\)的值．
 
 

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

參考答案

1. \(A\)
2. \(C\)
3. \(-1+\dfrac{\sqrt{3}}{3}\)
4. \(-\dfrac{1}{3}\)
5. \(\text { (1) } 1-\sqrt{2}\) \(\text { (2) } \sqrt{2}+1\)

【挑戰學霸】
解 設\(S=\sin ^{2} 1^{\circ}+\sin ^{2} 2^{\circ}+\sin ^{2} 3^{\circ}+\cdots+\sin ^{2} 89^{\circ}\) ①
又\(\because S=\sin ^{2} 89^{\circ}+\sin ^{2} 88^{\circ}+\sin ^{2} 87^{\circ}+\cdots+\sin ^{2} 1^{\circ}\)
\(=\cos ^{2} 1^{\circ}+\cos ^{2} 2^{\circ}+\cos ^{2} 3^{\circ}+\cdots+\cos ^{2} 89^{\circ}\) ②
由①+②得 \(2S=89\)，則\(S=\dfrac{89}{2}\)．