誘導公式練習題含答案答案：1.已知$tan(x+2)=5$，求$sinxcosx$。解：根據$tan(x+2)=\frac{sin(x+2)}{cos(x+2)}=5$，可得$sin(x+2)=5cos(x+2)$。將其代入$sinxcosx$中，得：$$sinxcosx=\frac{1}{2}(sin(x+2)+sin(x-2))cosx=\frac{5}{2}(cos^2(x+2)-cos^2(x-2))$$2.求$cos390°$。解：將$390°$轉化為弧度制，得$\frac{13\pi}{6}$。根據余弦的周期性，$cos(\frac{13\pi}{6})=cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$。3.已知$cos\frac{1}{2}\pi=\frac{1}{2}$，求$cos\frac{1}{3}\pi$。解：根據余弦的半角公式，$cos\frac{1}{3}\pi=\sqrt{\frac{1+cos\frac{2}{3}\pi}{2}}=\sqrt{\frac{1+\frac{1}{2}}{2}}=\frac{\sqrt{3}}{2}$。4.已知$sin^2x-4=10$，求$sin\alpha$。解：根據題意，$sin^2x=14$，因此$|sinx|=\sqrt{14}$。由于$sin\alpha$的取值范圍為$[-1,1]$，因此$sin\alpha=-\frac{\sqrt{14}}{2}$。5.已知$tan\alpha=3$，求$\frac{2cos\alpha-3sin\alpha}{2sin\alpha+cos\alpha}$。解：根據題意，$\frac{2cos\alpha-3sin\alpha}{2sin\alpha+cos\alpha}=\frac{2\frac{1}{\sqrt{10}}-3\frac{3}{\sqrt{10}}}{2\frac{3}{\sqrt{10}}+\frac{1}{\sqrt{10}}}=-\frac{3}{5}$。6.已知$sin(\alpha-\frac{1}{4}\pi)=\frac{3}{4}$，求$cos(\alpha+\frac{1}{4}\pi)$。解：根據正弦的差角公式，$sin(\alpha-\frac{1}{4}\pi)=sin\alphacos\frac{1}{4}\pi-cos\alphasin\frac{1}{4}\pi=\frac{\sqrt{2}}{2}(sin\alpha-cos\alpha)$。因此$sin\alpha-cos\alpha=\frac{3\sqrt{2}}{4}$。根據余弦的和角公式，$cos(\alpha+\frac{1}{4}\pi)=cos\alphacos\frac{1}{4}\pi+sin\alphasin\frac{1}{4}\pi=\frac{\sqrt{2}}{2}(sin\alpha+cos\alpha)$。因此$sin\alpha+cos\alpha=\frac{\sqrt{2}}{2}cos(\alpha-\frac{1}{4}\pi)=\frac{\sqrt{2}}{2}\times\frac{4}{5}=\frac{2\sqrt{2}}{5}$。將兩式相加，得$2cos\alpha=\frac{5\sqrt{2}}{4}$，因此$cos\alpha=\frac{5\sqrt{2}}{8}$。代入公式，得$cos(\alpha+\frac{1}{4}\pi)=\frac{\sqrt{2}}{2}(sin\alpha+cos\alpha)=\frac{3\sqrt{2}}{8}$。7.已知$cos\alpha=-\frac{5}{\sqrt{24}}$，且$\alpha$是第三象限角，求$tan\alpha$。解：根據三角函數的定義，$cos\alpha=-\frac{5}{\sqrt{24}}=\frac{-5}{2\sqrt{6}}$，$sin\alpha=-\sqrt{1-cos^2\alpha}=\frac{\sqrt{11}}{2\sqrt{6}}$。因此$tan\alpha=\frac{sin\alpha}{cos\alpha}=-\frac{\sqrt{11}}{5}$。8.已知$tan\alpha=\sqrt{3}$，且$\alpha$是第三象限角，求$cos\alpha-sin\alpha$。解：根據三角函數的定義，$tan\alpha=\sqrt{3}=\frac{sin\alpha}{cos\alpha}$，因此$cos\alpha=\frac{1}{2}$，$sin\alpha=-\frac{\sqrt{3}}{2}$。因此$cos\alpha-sin\alpha=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2}$。9.已知$f(\alpha)=\frac{cos(-\alpha)sin(-\pi-\alpha)}{sin(\frac{\pi}{2}-\alpha)cos(\frac{\pi}{3}-\alpha)}$。(1)化簡$f(\alpha)$。解：根據三角函數的性質，$cos(-\alpha)=cos\alpha$，$sin(-\pi-\alpha)=-sin\alpha$，$sin(\frac{\pi}{2}-\alpha)=cos\alpha$，$cos(\frac{\pi}{3}-\alpha)=\frac{1}{2}cos\alpha+\frac{\sqrt{3}}{2}sin\alpha$。因此$f(\alpha)=\frac{cos\alphasin\alpha}{cos\alpha(\frac{1}{2}cos\alpha+\frac{\sqrt{3}}{2}sin\alpha)}=\frac{2sin\alpha}{cos\alpha+\sqrt{3}sin\alpha}$。(2)若$\alpha$是第三象限角，且$sin(\alpha-\pi)=\frac{1}{\sqrt{6}}$，求$f(\alpha)$的值。解：根據題意，$sin\alpha=-\frac{1}{\sqrt{6}}$。代入$f(\alpha)$的式子，得$f(\alpha)=\frac{2\times(-\frac{1}{\sqrt{6}})}{-\frac{1}{\sqrt{6}}+\sqrt{3}\times(-\frac{1}{\sqrt{6}})}=\frac{2}{\sqrt{3}-1}$。10.在$\triangleABC$中，$\angleA,\angleC$均為銳角，且$|sinA|+(cosC-\frac{1}{\sqrt{2}})^2=\frac{1}{2}$，求$\angleB$的度數。解：根據題意，$|sinA|=\frac{1}{2}-(cosC-\frac{1}{\sqrt{2}})^2$。由于$\angleA,\angleC$均為銳角，因此$0<|sinA|\leqslant1$，即$\frac{1}{2}-(cosC-\frac{1}{\sqrt{2}})^2\leqslant1$。解得$cosC-\frac{1}{\sqrt{2}}\geqslant-\frac{1}{2}$。根據余弦的定義，$cosC=\frac{b^2+a^2-c^2}{2ab}$，因此$b^2+a^2-c^2\geqslant\frac{1}{\sqrt{2}}ab$。根據余弦定理，$cosB=\frac{a^2+c^2-b^2}{2ac}$，因此$cosB=\frac{a^2+c^2-b^2}{2ac}\leqslant\frac{a^2+c^2-\frac{1}{\sqrt{2}}ab}{2ac}=\frac{a}{2c}-\frac{1}{2\sqrt{2}}$。因此$\angleB$的度數不超過$60°$。11.已知$sin(30°+\alpha)=\frac{5}{3}$，$60°<\alpha<150°$，求$cos\alpha$。解：根據正弦的和角公式，$sin(30°+\alpha)=\frac{1}{2}cos\alpha+\frac{\sqrt{3}}{2}sin\alpha=\frac{5}{3}$。由于$60°<\alpha<150°$，因此$sin\alpha>0$，$cos\alpha<0$。因此$\frac{1}{2}|cos\alpha|<\frac{5}{6}$，$|\sqrt{3}sin\alpha|<\frac{5}{6}$。由于$\alpha$是第二象限角或第三象限角，因此$cos\alpha=-\sqrt{1-sin^2\alpha}=-\sqrt{1-\frac{25}{9}}=-\frac{4}{3}$。12.已知$f(x)=\frac{sin(\frac{\pi}{4}+x)-2cos(\pi+x)}{sin(\pi-x)+cos(-x)}$。(1)求$f(\frac{\pi}{4})$的值。解：將$x=\frac{\pi}{4}$代入$f(x)$的式子，得$f(\frac{\pi}{4})=\frac{sin(\frac{\pi}{2})-2cos(\frac{5}{4}\pi)}{cos\frac{\pi}{4}+sin\frac{\pi}{4}}=-\sqrt{2}$。(2)若$f(\alpha)=2$，$\alpha$是第三象限角，求$tan\alpha$和$sin\alpha$的值。解：根據$f(x)$的式子，可得：$$f(x)=\frac{\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx-2(-cosx)}{-sinx+cosx}=\frac{2sinx+\sqrt{2}cosx}{sinx-cosx}$$因此$2sin\alpha+\sqrt{2}cos\alpha=2cos\alpha-sin\alpha$，即$3sin\alpha=-3cos\alpha$。因此$tan\alpha=-1$，$sin\alpha=-\frac{1}{\sqrt{10}}$。13.已知$f(\alpha)=\frac{sin(\alpha-\frac{\pi}{3})cos(\frac{\pi}{2}+\alpha)}{sin(\alpha-\frac{\pi}{2})+cos(-\alpha)}$。(1)化簡$f(\alpha)$。解：根據三角函數的性質，$sin(\alpha-\frac{\pi}{3})=\frac{1}{2}sin\alpha-\frac{\sqrt{3}}{2}cos\alpha$，$cos(\frac{\pi}{2}+\alpha)=-sin\alpha$，$sin(\alpha-\frac{\pi}{2})=-cos\alpha$，$cos(-\alpha)=cos\alpha$。因此$f(\alpha)=\frac{(\frac{1}{2}sin\alpha-\frac{\sqrt{3}}{2}cos\alpha)(-sin\alpha)}{-cos\alpha+cos\alpha}=\sqrt{3}tan\alpha$。(2)若$sin(\alpha+\frac{\pi}{2})=-\sqrt{6}$，求$f(\alpha+\pi)$的值。解：根據$f(\alpha)$的式子，可得$f(\alpha+\pi)=\sqrt{3}tan(\alpha+\pi)=\sqrt{3}tan\alpha=-\sqrt{3}\sqrt{6}=-3\sqrt{2}$。14.已知$f(x)=\frac{sin(\frac{\pi}{2}+x)-2cos(\pi+x)}{sin(\pi-x)+cos(-x)}$。(1)求$f(\frac{\pi}{4})$的值。解：將$x=\frac{\pi}{4}$代入$f(x)$的式子，得$f(\frac{\pi}{4})=\frac{cos\frac{3}{4}\pi-2cos\frac{5}{4}\pi}{-cos\frac{\pi}{4}-sin\frac{\pi}{4}}=-\sqrt{2}$。(2)若$f(\alpha)=1$，求$tan\alpha$和$sin\alpha$的值。解：根據$f(x)$的式子，可得：$$f(x)=\frac{\frac{\sqrt{2}}{2}cosx-\frac{\sqrt{2}}{2}sinx-2(-cosx)}{-sinx+cosx}=\frac{2sinx+\sqrt{2}cosx}{cosx-sinx}$$因此$2sin\alpha+\sqrt{2}cos\alpha=cos\alpha-sin\alpha$，即$3sin\alpha=-\sqrt{sin(?????)cos(2?????)cos(2???)??(??)=?cos??解法一：利用誘導公式化簡：sin(?????)cos(2?????)cos(2???)=-sin??sin2??=-sin??(1-cos2??)=-sin??+sin??cos2??代入原式得：??(??)=-cos??+sin??cos2??解法二：由于??是第三象限角，且sin(??-??)=-1/5，可得sin??=-1/5，利用同角三角函數間的基本關系求出cos??的值：cos??=√(1-sin2??)=√(1-1/25)=√24/5將cos??代入原式，利用誘導公式化簡即可得到結果：??(??)=-cos??+sin??cos2??=-√24/5-1/5*(-24/5)=-2√6/5綜上所述，答案為B.cos(4)+2cos(4)sin(4)+cos(4)=3tan(4)+1=3×1+1=4.(2)∵sin(?????)=sin??,cos(???)=cos??∴??(?????)=cos(?????)+2cos(?????)sin(?????)+cos(?????)=?cos??+2(?cos??)sin???sin??=?3cos??.∴??(??)+??(?????)=?cos??+3cos??=2cos??∴??(??)???(?????)=tan???3tan??=?2tan??∴??(??)=12(??(??)+??(?????))+12(??(??)???(?????))=12(2cos??)?12(?2tan??)=cos??+tan??.【考點】三角函數的基本性質【解析】(1)直接帶入公式計算即可；(2)利用奇偶性和三角函數的基本關系式，將??(?????)表示為??(??)的形式，再利用加減法公式進行計算。【解答】(1)∵??(??)=cos??+2cos??sin??+cos??3tan??+1sin(+??)?2cos(??+??)sin(?????)+cos(???)??23443√3+25×2=13?4√3．10,??tan+14∴??(4)=4.(2)∵sin(?????)=sin??,cos(???)=cos??∴??(?????)=cos(??