第 12 课 NCERT 解决方案 - 数学第一部分 - 第 2 章反三角函数 - 第 2 章杂项练习 |设置 1

问题 1. 找出${\cos }^{-1}(\cos \frac {13\pi} {6})$

解决方案：

We know that $\cos^{-1} (\cos x)=x$ 

Here, $\frac {13\pi} {6} \notin [0,\pi].$

Now, ${\cos }^{-1}(\cos \frac {13\pi} {6})$ can be written as :

${\cos }^{-1}(\cos \frac {13\pi} {6})={\cos }^{-1}[\cos( 2\pi+\frac {\pi} {6})]$, where $\frac{\pi} {6} \in [0,\pi].$

Hence, the value of ${\cos }^{-1}(\cos \frac {13\pi} {6})$ = π/6

问题 2. 找出$\tan^{-1}(\tan \frac {7\pi}{6})$

解决方案：

We know that $\tan^{-1} (\tan x)=x$

Here, $\frac {7\pi}{6} \notin(\frac{-\pi}{2},\frac{\pi}{2})$

Now, $\tan^{-1}(\tan \frac {7\pi}{6})$ can be written as:

$\tan^{-1}(\tan \frac {7\pi}{6})=\tan^{-1}[\tan( 2\pi -\frac {5\pi}{6})]$      $-[\tan(2\pi-x)=-\tan x]$

$\tan^{-1}[-\tan(\frac {5\pi}{6}) ]=\tan^{-1}[\tan(-\frac {5\pi}{6})]$

$=\tan^{-1}[\tan(\pi-\frac {5\pi}{6})]=\tan^{-1}[\tan( \frac {\pi}{6})],$ where $\frac{\pi}{6} \in (\frac{-\pi}{2},\frac{\pi}{2})$

Hence, the value of $\tan^{-1}(\tan\frac{7\pi}{6})$ = π/6

问题 3. 证明$2\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{24}{7}$

解决方案：

Let $\sin^{-1} \frac{3}{5}=x$          -(1)

sin x = 3/5

So,$\cos x = \sqrt{1-(\frac{3}{5})^2 }$= 4/5

tan x = 3/4

Hence, $x=\tan^{-1} \frac{3}{4}$

Now put the value of x from eq(1), we get

$\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}$

Now, we have

L.H.S $= 2 \sin^{-1} \frac{3}{5}=2 \tan^{-1} \frac{3}{4}$

= $\tan^{-1}(\frac{2 \times \frac{3}{4}}{1-(\frac{3}{4})^{2}})$                –$[2\tan^{-1} x=\tan^{-1} \frac{2x}{1-x^2}]$

$= \tan^{-1}(\frac{ \frac{3}{2}}{\frac{16-9}{16}})=\tan^{-1} (\frac{3}{2} \times \frac{16}{7})$

$=\tan^{-1} \frac{24}{7}$

Hence, proved.

问题 4. 证明$\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{77}{36}$

解决方案：

Let $\sin^{-1} \frac{8}{17}=x$ 

Then sin x = 8/17 

 cos x =$\sqrt{1-(\frac{8}{17})^2}=\sqrt \frac{225}{289}$ = 15/17

Therefore, $\tan x=\frac{8}{15}\implies x=\tan^{-1}\frac{8}{15}$

$\sin^{-1} \frac{8}{17}=\tan^{-1} \frac{8}{17}$         -(1)

Now, let $\sin^{-1} \frac{3}{5}=y$

Then, sin y = 3/5 

$\cos y=\sqrt{1- (\frac{3}{5})^2}=\sqrt{ (\frac{16}{25})}$ = 4/5

$\therefore \tan y =\frac{3}{4} \implies y=\tan^{-1} \frac{3}{4}$

$\therefore \sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}$     -(2)

Now, we have:

L.H.S.$=\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}$

 From equation(1) and (2), we get

= $\tan^{-1} \frac{8}{15}+\tan^{-1} \frac{3}{4}$                         

= $\tan^{-1} \frac{{\frac{8}{15}+ \frac{3}{4}} }{1-{\frac{8}{15}\times \frac{3}{4}}}$

= $\tan^{-1}(\frac{32+45}{60-24})$                                 –$[\tan^{-1} x + \tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$

= $\tan^{-1} \frac{77}{36}$

Hence proved

问题 5. 证明$\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

解决方案：

Let $\cos^{-1}\frac{4}{5}= x$

Then, cos x = 4/5

$\sin x = \sqrt {1- (\frac{4}{5})^{2}}$ = 3/5

$\therefore \tan x =\frac{3}{4} \implies x=\tan^{-1} \frac{3}{4}$

$\therefore \cos^{-1} \frac{4}{5}=\tan^{-1} \frac{3}{4}$        -(1)

Now let $\cos^{-1} \frac{12}{13}=x$

Then, cos y = 3/4 

$\sin^{-1} y=\frac{5}{13}$

$\therefore\tan y= \frac{5}{12} \implies y=\tan^{-1} \frac{5}{12}$

$\therefore \cos ^{-1} \frac{12}{13}=\tan^{-1} \frac{5}{12}$            -(2)

Let $\cos^{-1} \frac{33}{65}=z$ 

Then, cos z = 33/65

sin z = 56/65 

$\therefore \tan z = \frac{56}{65} \implies z= tan^{-1}\frac{56}{33}$

$\therefore \cos^{-1} \frac{33}{65}= \tan^{-1} \frac{56}{33}$    -(3)

Now, we will prove that : 

L.H.S. $=\cos^{-1} \frac{3}{5}\cos^{-1} \frac{12}{13}$

From equation (1) and equation (2)

= $\tan^{-1} \frac{3}{4}+\tan^{-1} \frac{5}{12}$                                                                 

= $\tan^{-1} \frac{{\frac{3}{4}+ \frac{5}{12}} }{1-{\frac{3}{4}\times \frac{5}{12}}}$         –$[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$

= $\tan^{-1} \frac{36+20}{48-15}$

= $\tan^{-1} \frac{56}{33}$

Using equation(3)

= $\tan^{-1} \frac{56}{33}$                                                                                                  

Hence proved

问题 6. 证明$\cos^{-1} \frac{12}{13}+\sin^{-1} \frac{3}{5}=\sin^{-1} \frac{56}{65}$

解决方案：

Let $\sin^{-1} \frac{3}{5}=x$  

Then, sin x = 3/5 

$\cos x =\sqrt{1- (\frac{3}{5})^{2}}=\sqrt \frac{16}{25}$ = 4/5

$\therefore \tan x = \frac{3}{4} \implies x= \tan^{-1} \frac{3}{4}$

$\therefore \sin^{-1} \frac{3}{5}= \tan^{-1} \frac{3}{4}$           -(1)

Now, let $\cos^{-1} \frac{12}{13}=y$  

Then, cos y = 12/13 and sin y = 5/13

$\therefore \tan y = \frac{5}{12} \implies y= \tan^{-1} \frac{5}{12}$

$\therefore \cos^{-1} \frac{12}{13}= \tan^{-1} \frac{5}{12}$        -(2)                                                             

Let $\sin^{-1} \frac{56}{65}=z$

Then, sin z = 56/65 and cos z = 33/65

$\therefore \tan z = \frac{56}{33} \implies z=\tan ^{-1} \frac{56}{33}$

$\therefore \sin^{-1} \frac{56}{65}= \tan^{-1} \frac{56}{33}$           -(3)

Now, we have:

L.H.S.=$\cos^{-1} \frac{12}{13}+ \sin^{-1} \frac{3}{5}$

From equation(1) and equation(2)

=$\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{3}{4}$                                                                           

= $\tan^{-1} \frac{{\frac{5}{12}+ \frac{3}{4}} }{1-{\frac{5}{12}\times \frac{3}{4}}}$               –$[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$

= $\tan^{-1} \frac{20+36}{48-15}$

= $\tan^{-1} \frac{56}{33}$

From equation (3)

= $\sin^{-1} \frac{56}{65}$    

Hence proved                                                                                                           

问题 7. 证明$\tan^{-1} \frac{63}{16}= \sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}$

解决方案：

Let $\sin^{-1} \frac{5}{13}=x$   

Then, sin x = 5/13 and cos x = 12/13.

$\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}$

$\therefore \tan x= \frac{5}{12} \to x= \tan^{-1} \frac{5}{12}$

$\therefore \sin^{-1} \frac{5}{13}= \tan^{-1} \frac{5}{12}$                   -(1)

Let $\cos^{-1} \frac{3}{5}=y$     

Then, cos y = 3/5 and sin y = 4/5

$\therefore \tan y= \frac{4}{3} \implies y= \tan^{-1}\frac{4}{3}$

$\therefore \cos ^{-1}\frac{3}{5}=\tan^{-1} \frac{4}{3}$                -(2)

From equation(1) and (2), we have

R.H.S.$=\sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}$

=$\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{4}{3}$

= $\tan^{-1} \frac{{\frac{5}{12}+ \frac{4}{3}} }{1-{\frac{5}{12}\times \frac{4}{3}}}$          –$[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$

=$\tan^{-1} \frac{15+48}{36-20}$

=$\tan^{-1} \frac{63}{16}$

L.H.S = R.H.S

Hence proved

问题 8. 证明$\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}=\frac{\pi}{4}$

解决方案：

L.H.S.$=\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}$

= $\tan^{-1} \frac{{\frac{1}{5}+ \frac{1}{7}} }{1-{\frac{1}{5}\times \frac{1}{7}}} +\tan^{-1} \frac{{\frac{1}{3}+ \frac{1}{8}} }{1-{\frac{1}{3}\times \frac{1}{8}}}$     –$[\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}]$

= $\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}$

= $\tan^{-1} \frac{12}{34}+\tan^{-1} \frac{11}{23}$

= $\tan^{-1} \frac{6}{17}+\tan^{-1} \frac{11}{23}$

= $\tan^{-1} \frac{{\frac{6}{17}+ \frac{11}{23}} }{1-{\frac{6}{17}\times \frac{11}{23}}}$

= $\tan^{-1} \frac{138 + 187}{391-66}$

= $\tan^{-1} \frac{325}{325}=\tan^{-1} 1$

= π/4 

L.H.S = R.H.S

Hence proved

问题 9. 证明$\tan^{-1} \sqrt x= \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x}),x\in[0,1]$

解决方案：

Let x = tan²θ

Then,$\sqrt x=\tan \theta \implies\theta=\tan^{-1} \sqrt x.$

$\therefore \frac{1-x}{1+x}+\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta$

Now, we have

R.H.S = $\frac{1}{2} \cos ^{-1}(\frac{1-x}{1+x})= \frac{1}{2} \cos ^{-1} (\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan^{-1}\sqrt x$

L.H.S = R.H.S

Hence proved

问题 10. 证明$\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})=\frac{x}{2},x\in(0,\frac{\pi}{4})$

解决方案：

Consider$(\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})$

By rationalizing

=$\frac{(\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x}))^{2}}{(\sqrt{ ({1+ \sin x})}-\sqrt({1- \sin x}))^{2}}$                         

=$\frac{( {1+ \sin x)} + {(1-\sin x)} + 2 \sqrt{(1+\sin x)(1-\sin x)}}{{ {1+ \sin x}}-{1+\sin x}}$

=$\frac{2(1+\sqrt{1-\sin^{2}})}{2\sin x}=\frac{1+\cos x}{\sin x}=\frac{2\cos ^{2}\frac{x}{2}}{2\sin \frac{x}{2}\cos\frac{x}{2}}$

= $\cot \frac{x}{2}$

L.H.S = $\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})={\cot}^{-1}({\cot( \frac x 2)})$= x/2

L.H.S = R.H.S

Hence proved

第 2 章反三角函数 - 第 2 章杂项练习 |设置 2