任意角、三角函数的概念

答案

$$答案：(1)D,(2)C;提示：\alpha \in \left ( \frac{\pi }{2}+2k\pi , \pi +2k \pi \right ) ,k\in{Z},则\frac{\alpha }{2} \in \left ( \frac{\pi }{4}+k\pi , \frac{\pi }{2} +k \pi \right ) ,~\\k\in{Z}，在一、三象限;\frac{\alpha }{3} \in \left ( \frac{\pi }{6}+\frac{2k\pi}{3} , \frac{\pi }{3} +\frac{2k\pi}{3} \right ) ,k\in{Z}，在第一或第二或第四象限.$$

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答案

$$答案：D;提示：x为第一象限角，原理同上~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:-2\tan \alpha;提示：原式\Rightarrow \sqrt{\frac{(1+\sin \alpha)^2}{1-\sin^2 \alpha}}-\sqrt{\frac{(1-\sin \alpha)^2}{1-\sin^2 \alpha}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案：B;提示： P(\sin \alpha , \boldsymbol{\operatorname { t a n }} \alpha )在第四象限,\sin \alpha >0且 \tan {\alpha}<0,\therefore \alpha在第二象限.~~~~~~~$$

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答案

$$答案：a>0, 可令a=1,\sin \alpha=\frac{4}{5}, \cos \alpha=\frac{3}{5}, \tan \alpha=\frac{4}{3};~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\a<0, 可令a=-1,\sin \alpha=-\frac{4}{5}, \cos \alpha=-\frac{3}{5}, \tan \alpha=\frac{4}{3} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案: C; 提示： 由题意得点 P(-8 m,-3) , r=\sqrt{64 m^{2}+9} 所以 \cos \alpha=\frac{-8 m}{\sqrt{64 m^{2}+9}}~~~~~~\\=-\frac{4}{5} , 所以 m>0 , 解得 m=\frac{1}{2} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案：\frac{\sqrt{17}}{34} 或 -\frac{\sqrt{17}}{34} ;提示:在第一象限取点(1,4),得 \cos \alpha=\frac{1}{\sqrt{17}}, \sin \alpha=\frac{4}{\sqrt{17}}, ~~~~~~~\\\tan \alpha=4 ;在第三象限取点(-1,-4),得 \cos \alpha=-\frac{1}{\sqrt{17}}, \sin \alpha=-\frac{4}{\sqrt{17}}, \tan \alpha=4 .~~~~~~~~~\\★注：由y=4x,\tan \alpha=4,\sin \alpha和\cos \alpha不唯一.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:(1)\alpha为第三象限角时， \cos \alpha=-\frac{4}{5}, \tan \alpha =\frac{3}{4};\alpha为第四象限角时，~~~~~~~~~~~~~~~~~~~~~\\ \cos \alpha=\frac{4}{5}, \tan \alpha =-\frac{3}{4}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\(2)\alpha为第二象限角时， \sin \alpha=\frac{\sqrt{3}}{2}, \cos \alpha =-\frac{1}{2};\alpha为第四象限角时，~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \sin \alpha=-\frac{\sqrt{3}}{2}, \cos \alpha =\frac{1}{2}.★注：可令y=\sqrt{3},x=1,快速求|\sin x|和|\cos x|,再加符号.\\(3)-\frac{\sqrt{2} }{2} ;-\frac{\sqrt{3} }{2};\frac{\sqrt{3} }{2};-\frac{\sqrt{3} }{2}.提示:根据对称性转化到第一象限角.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:C;提示:y_1+x_2=\sin{(\frac{\pi}{2}+2t)}+2\cos t.~~~~~~~~~~~~~~$$

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三角公式的应用

答案

$$答案:(1)3;(2)\frac{2}{5};(3)1;提示：利用齐次式，转化为\tan \alpha.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案: -\frac{3}{5} ; 提示： 由已知 \Rightarrow \tan \left(\alpha-\frac{\pi}{4}\right)=\frac{1}{3}, \Rightarrow \tan \alpha=2, \therefore \cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha\\= \frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}=-\frac{3}{5} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案：\frac{1}{2} ;提示：\tan \alpha=\frac{1}{3}, \Rightarrow \sin2 \alpha=\frac{3}{5},\cos2 \alpha=\frac{4}{5},原式~2 \sin \beta=\sin (2 \alpha+\beta)~~~~~~~~\\ 展开代入得 \tan \beta=\frac{1}{2}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案 :B;提示： \tan \alpha=\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{2 \cos ^{2} \alpha}=\frac{ \tan ^{2} \alpha+1}{2} ,即 \tan ^{2} \alpha-2 \tan \alpha+1=0 , ~~~~~~\\解得 \tan \alpha=1 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:(1) \sin \theta-\cos \theta=\frac{7}{5};(2) \sin \theta=\frac{4}{5} ; (3)\cos \theta=-\frac{3}{5} ; (4)\tan \theta=-\frac{4}{3} .~~~~~~~~~~~~~~~~~~~~~~~\\提示：由题意知 \sin \theta+\cos \theta=\frac{1}{5}, 平方得 2 \sin \theta \cos \theta=-\frac{24}{25}<0 , 又 \because \theta \in(0, \pi), ~~~~~~~~~\\\therefore \frac{\pi}{2}< \theta<\pi, \quad \therefore \sin \theta-\cos \theta>0, \quad \therefore \sin \theta-\cos \theta=\sqrt{1-2 \sin \theta \cos \theta}=\frac{7}{5}, ~~~~~~~~~~~~~~\\与已知联立， \sin \theta=\frac{4}{5}, \cos \theta=-\frac{3}{5} . \therefore \tan \theta=-\frac{4}{3}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案： -\frac{24}{175} ;提示： 由已知\Rightarrow \sin x+\cos x=\frac{1}{5}, \Rightarrow 2 \sin x \cos x=-\frac{24}{25} ,\Rightarrow~~~~~~~~~~~~~~~~~\\(\sin x-\cos x)^{2}=1-2 \sin x \cos x =\frac{49}{25} , 缩小角的范围\Rightarrow -\frac{\pi}{2}< x< 0 \Rightarrow ~~~~~~~~~~~~~~~~~~~~~~\\\sin x-\cos x=-\frac{7}{5} ,\Rightarrow\sin x =-\frac{3}{5},\cos x=\frac{4}{5},\tan x=-\frac{3}{4}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:-\tan \alpha;提示：化简为\frac{(-\sin \alpha)(-\cos \alpha)(- \sin\alpha) (-\sin \alpha)}{(-\cos \alpha)\sin \alpha\sin \alpha\cos \alpha}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案 (1)4; 提示:原式 =\frac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}= \frac{4 \sin \left(30^{\circ}-10^{\circ}\right)}{\sin 20^{\circ}}=4 ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

$$(2)-1; 提示：\sin 40^{\circ} \cdot\left(\frac{\sin 10^{\circ}}{\cos 10^{\circ}}-\sqrt{3}\right)=\sin 40^{\circ} \cdot \frac{\sin 10^{\circ}-\sqrt{3} \cos 10^{\circ}}{\cos 10^{\circ}} ~~~~~~~~~~~~~~~~~~~~~~~~~\\ =\sin 40^{\circ} \cdot \frac{-2 \sin 50^{\circ}}{\cos 10^{\circ}}=\frac{-2 \sin 40^{\circ} \cdot \cos 40^{\circ}}{\cos 10^{\circ}}=\frac{-\sin 80^{\circ}}{\cos 10^{\circ}}=-1 ;$$

$$(3)\sqrt{3}; 提示:原式 =\frac{2 \cos \left(30^{\circ}+28^{\circ}\right)+\sin 28^{\circ}}{\cos 28^{\circ}}=\frac{\sqrt{3} \cos 28^{\circ}}{\cos 28^{\circ}}=\sqrt{3} ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

$$(4)-1;原式=\frac{\displaystyle\frac{1-\cos 70^{\circ}}{2}-\frac{1}{2}}{\cos 10^{\circ} \cdot \sin 10^{\circ}} =-\frac{\cos 70^{\circ}}{2 \sin 10^{\circ} \cdot \cos 10^{\circ}}=-\frac{\sin 20^{\circ}}{\sin 20^{\circ}}=-1 .~~~~~~~~~~~~~~~~~~~~~~~~$$

$$(5)2 \sin 2;原式=2 \sqrt{(\sin 2+\cos 2)^{2}}+\sqrt{4 \cos ^{2} 2} =2|\sin 2+\cos 2|+2|\cos 2|~~~~~~~~~~~~~~~~~\\=2(\sin 2+\cos 2)-2 \cos 2=2 \sin 2.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$(1)答案:2;提示： \boldsymbol{\operatorname { t a n }}(\alpha+\beta)=\frac{\boldsymbol{\operatorname { t a n }} \alpha+\tan \beta}{1-\boldsymbol{\operatorname { t a n }} \alpha \boldsymbol{\operatorname { t a n }} \beta}=1 , \therefore 1-\tan \alpha \boldsymbol{\operatorname { t a n }} \beta=\tan \alpha+\tan \beta , \\则 (1+\tan \alpha) \cdot(1+\tan \beta)=1+\tan \alpha+\tan \beta+\tan \alpha \boldsymbol{\operatorname { t a n }} \beta=2 .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\(2)答案:4;提示:同(1),\left(1+\tan 13^{\circ}\right)\left(1+\tan 32^{\circ}\right)=\left(1+\tan 17^{\circ}\right)\left(1+\tan 28^{\circ}\right)=2 .~~$$

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答案

$$答案:(1) f(x)=2 \sin \left(2\omega x-\frac{\pi}{6}\right);~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)f(x)=\sin \left(2 x-\frac{\pi}{3}\right)+\frac{1-\sqrt{3}}{2};~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (3) f(x)=\sqrt{3} \sin \left(2 \omega x-\frac{\pi}{3}\right) ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (4)f(x)=\frac{1}{2} \sin \left(2 x+\frac{\pi}{6}\right)+\frac{1}{2} ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (5)f(x)=\frac{1}{2} \sin \left(2 \omega x+\frac{\pi}{6}\right)-\frac{1}{4};~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (6)f(x)=\cos 2 x-\sin 2 x=-\sqrt{2} \sin \left(2 x-\frac{\pi}{4}\right);~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (7)f(x)=\sin \left(2 x-\frac{\pi}{6}\right)+\frac{1}{2}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

$$(1)提示: 降幂公式 \sin ^{2} \omega x=\frac{1-\cos 2 \omega x}{2} \Rightarrow f(x)=-\cos 2 \omega x+ \sqrt{3} \sin 2 \omega x~~~~~~~~~~~~~~~~~\\=2 \sin \left(2 \omega x-\frac{\pi}{6}\right) ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (2)提示:f(x) =(1-\sqrt{3}) \frac{1+\cos 2 x}{2}+\frac{1}{2} \sin 2 x+\frac{1}{2}\left(\sin ^{2} x-\cos ^{2} x\right) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ =\frac{1-\sqrt{3}}{2}-\frac{\sqrt{3} \cos 2 x}{2}+\frac{1}{2} \sin 2 x=\sin \left(2 x-\frac{\pi}{3}\right)+\frac{1-\sqrt{3}}{2};~~~~~~~~\\ (3) 提示:f(x) =\frac{\sqrt{3}}{2} \sin 2 \omega x-\frac{1}{2} \cos 2 \omega x-\cos 2 \omega x =\sqrt{3}\left(\frac{1}{2} \sin 2 \omega x-\frac{\sqrt{3}}{2} \cos 2 \omega x\right)~\\=\sqrt{3} \sin \left(2 \omega x-\frac{\pi}{3}\right);~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ \\ (4)提示: f(x)=\cos ^{2}\left(x+\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} \sin 2 x=\frac{1+\cos \left(2 x+\frac{\pi}{3}\right)}{2}+\frac{\sqrt{3}}{2} \sin 2 x ~~~~~~~~~~~~~~~~~~\\ =\frac{1}{2}+\frac{1}{4} \cos 2 x+\frac{\sqrt{3}}{4} \sin 2 x=\frac{1}{2} \sin \left(2 x+\frac{\pi}{6}\right)+\frac{1}{2} ;~~~~~~~~~~~~~~~~~~~~~~~\\ (5)提示: f(x)=\sin \omega x\left(\sin \omega x \cos \frac{\pi}{3}+\cos \omega x \sin \frac{\pi}{3}\right)-\sin ^{2} \omega x=\frac{\sqrt{3}}{2} \cos \omega x \sin \omega x-~\\\frac{1}{2} \sin ^{2} \omega x =\frac{\sqrt{3}}{4} \sin 2 \omega x-\frac{1}{4}(1-\cos 2 \omega x)=\frac{1}{2} \sin \left(2 \omega x+\frac{\pi}{6}\right)-\frac{1}{4}; \\ (6)提示: f(x)=\cos ^{4} x-\sin ^{4} x-2 \sin x \cdot \cos x=\left(\cos ^{2} x-\sin ^{2} x\right)\left(\cos ^{2} x+\sin ^{2} x\right)-~~\\\sin 2 x =\cos 2 x-\sin 2 x=-\sqrt{2} \sin \left(2 x-\frac{\pi}{4}\right);~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ (7)提示: f(x)=\frac{1+\cos 2 x}{2}+\frac{\sqrt{3}}{2} \sin 2 x-\cos 2 x=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2} \cos 2 x+\frac{1}{2} ~~~~~~~~~~~~~~~\\ =\sin \left(2 x-\frac{\pi}{6}\right)+\frac{1}{2}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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用已知角表示所求角

答案

$$答案：(1)\frac{5 \sqrt{3}+12}{26} ;提示： \cos \alpha=\cos \left[\left(\alpha+\frac{\pi}{6}\right)-\frac{\pi}{6}\right],其中，\sin \left(\alpha+\frac{\pi}{6}\right)=\frac{12}{13} .~~~$$

$$(2)答案:\frac{1-2\sqrt{6}}{6};提示: \cos \theta=\cos \left[\left(\theta+\frac{\pi}{6}\right)-\frac{\pi}{6}\right],其中，\because \theta\in(0,\pi),\therefore \theta+\frac{\pi}{6}\in\\ (\frac{\pi}{6},\frac{7\pi}{6})\sin\frac{\pi}{6}=\frac{1}{2}>\sin (\theta +\frac{\pi}{6})=\frac{1}{3},\therefore \theta +\frac{\pi}{6}不能是锐角,\cos(\theta +\frac{\pi}{6})=-\frac{2\sqrt{2}}{3}.~~~~~~~~~~$$

$$(3)答案:\frac{3-8\sqrt{2}}{15};提示:\sin B =\frac{4}{5},\cos B=\frac{3}{5},\sin B >\sin A,\therefore B > A,\cos A=\frac{2\sqrt{2}}{3}.\\ \sin (2 B+C) =\sin (\pi+B-A)=-\sin (B-A).~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:C;提示: \sin \beta=\sin [\alpha+(\beta-\alpha)]=\sin \alpha \cdot \cos (\beta-\alpha)+\cos \alpha \cdot \sin (\beta-\alpha), \alpha, \beta ~~ \\均为锐角, 所以 \cos \alpha=\frac{\sqrt{5}}{5}, \cos (\beta-\alpha)=\frac{3 \sqrt{10}}{10} .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案：-\frac{4}{5} ;提示： \cos \left(\alpha+\frac{\pi}{4}\right)=\cos \left[(\alpha+\beta)-\left(\beta-\frac{\pi}{4}\right)\right],其中,\because \alpha+\beta \in~~~~~~~~~~~~~~\\ \left(\frac{3 \pi}{2}, 2 \pi\right) , \therefore \cos (\alpha+\beta)=\frac{4}{5}; \because \beta-\frac{\pi}{4} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right) ,\therefore \cos \left(\beta-\frac{\pi}{4}\right)=-\frac{7}{25} .~~~~~~~~~~~~~~~~~$$

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答案

$$答案：\frac{4-3 \sqrt{3}}{10} ;提示:令t=\theta+\frac{\pi}{4},则\cos t=\frac{\sqrt{10}}{10}且t \in \left ( \frac{\pi}{4}, \frac{3\pi}{4} \right ) ,\theta =t- \frac{\pi}{4}, ~~~~~~~~~~~~\\则\sin \left(2 \theta-\frac{\pi}{3}\right)= \sin \left(2 t-\frac{2\pi}{3}\right),其中\sin 2t=\frac{3}{5},\cos 2t=-\frac{4}{5}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案:\frac{7\sqrt{2} }{50} ;提示：令\alpha +\frac{\pi }{6} =t,则\cos t=\frac{4}{5} ,\sin t=\frac{3}{5}, \sin \left(2 \alpha+\frac{\pi}{12}\right)=\sin (2t-\frac{\pi }{4} ) ~~\\=\frac{\sqrt{2} }{2} (\sin 2t-\cos 2t)=\frac{7\sqrt{2} }{50}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

答案

$$答案:C;令\frac{\pi}{3}-2 x=t,\cos t=-\frac{7}{8}, \sin \left(x+\frac{\pi}{3}\right) =\cos \frac{t}{2}=\pm \frac{1}{4}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

答案

$$答案: \frac{4 \sqrt{2}}{3} ; 提示：令 \frac{\pi}{3}-x=t, \therefore \frac{\pi}{6} < t<\frac{\pi}{3}, 且 x=\frac{\pi}{3}-t\therefore \sin t=\frac{1}{3} ，方法同上.~~~~$$

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答案

$$答案: (1)[-\sqrt3,2\sqrt3];(2)[-4,5]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:(1)f(x)=2\sqrt{3}\sin(x + \frac{\pi}{6}),x\in[0,\pi],x+\frac{\pi}{6}\in[\frac{\pi}{6},\frac{7\pi}{6}], f(x)值域为[-\sqrt3,2\sqrt3].\\ (2)f(x)=5(\frac{3}{5}\sin x+\frac{4}{5}\cos x)，设\cos\varphi=\frac{3}{5}，\sin\varphi=\frac{4}{5}，\varphi\in(0,\frac{\pi}{2}),~~~~~~~~~~~~~~~~~~~~~~~~\\ \therefore f(x)=5\sin(x + \varphi)，x+\varphi\in[\varphi,\pi+\varphi]，f(x)_{\max}=5，~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ f(x)_{\min}=5\sin(\pi+\varphi)= - 5\sin\varphi=-4, \therefore f(x)值域[-4,5]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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答案

$$答案: (1)[-\sqrt3,2\sqrt3];(2)[-4,5]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 提示:(1)f(x)=2\sqrt{3}\sin(x + \frac{\pi}{6}),x\in[0,\pi],x+\frac{\pi}{6}\in[\frac{\pi}{6},\frac{7\pi}{6}], f(x)值域为[-\sqrt3,2\sqrt3].\\ (2)f(x)=5(\frac{3}{5}\sin x+\frac{4}{5}\cos x)，设\cos\varphi=\frac{3}{5}，\sin\varphi=\frac{4}{5}，\varphi\in(0,\frac{\pi}{2}),~~~~~~~~~~~~~~~~~~~~~~~~\\ \therefore f(x)=5\sin(x + \varphi)，x+\varphi\in[\varphi,\pi+\varphi]，f(x)_{\max}=5，~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ f(x)_{\min}=5\sin(\pi+\varphi)= - 5\sin\varphi=-4, \therefore f(x)值域[-4,5]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

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