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2 arctan(x) = arctan[2x/(1−x^2)]


Kita akan belajar bagaimana membuktikan sifat invers fungsi trigonometri,  $2arctan(x)=arctan\left ( \frac{2x}{1-x^2} \right )=arcsin\left ( \frac{2x}{1+x^2} \right )=arccos\left ( \frac{1-x^2}{1+x^2} \right ) $

atau,  $2tan^{-1}x=tan^{-1}\left ( \frac{2x}{1-x^2} \right )=sin^{-1}\left ( \frac{2x}{1+x^2} \right )=cos^{-1}\left ( \frac{1-x^2}{1+x^2} \right ) $ 

Bukti:

Misal, tan−1 x = θ

Oleh karena itu, tan θ = x

Kita tahu bahwa,

$tan2\theta =\frac{2tan\theta }{1-tan^{2}\theta }$

 $tan2\theta =\frac{2x }{1-x^{2} }$

 $ 2\theta =tan^{-1}\frac{2x }{1-x^{2} }$ $ 2tan^{-1}x=tan^{-1}\frac{2x }{1-x^{2} }$.....................(1)  

Selanjutnya

$sin2\theta =\frac{2tan\theta }{1+tan^{2}\theta }$

 $sin2\theta =\frac{2x }{1+x^{2} }$

 $ 2\theta =sin^{-1}\frac{2x }{1-x^{2} }$ $ 2tan^{-1}x=sin^{-1}\frac{2x }{1+x^{2} }$.....................(2)  


Sekarang

$cos2\theta =\frac{1-tan^{2}\theta }{1+tan^{2}\theta }$

 $cos2\theta =\frac{1-x^{2} }{1+x^{2} }$

 $ 2\theta =cos^{-1}\frac{1-x^{2} }{1+x^{2} }$ $ 2tan^{-1}x=cos^{-1}\frac{1-x^{2} }{1+x^{2} }$.....................(3)  

oleh karena itu, dari (i), (ii) dan (iii) kita dapatkan,

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


atau $2arctan(x)=arctan\left ( \frac{2x}{1-x^2} \right )=arcsin\left ( \frac{2x}{1+x^2} \right )=arccos\left ( \frac{1-x^2}{1+x^2} \right ) $

Contoh penyelesaian tentang sifat fungsi invers $2arctan(x)=arctan\left ( \frac{2x}{1-x^2} \right )=arcsin\left ( \frac{2x}{1+x^2} \right )=arccos\left ( \frac{1-x^2}{1+x^2} \right ) $

Contoh 1
Tentukan nilai fungsi invers $tan(2tan^{-1}{\frac{1}{5}})$

Jawab:
$tan(2tan^{-1}{\frac{1}{5}})$ 

$=tan\left (tan^{-1}\frac{2\times \frac{1}{5}}{1-(\frac{1}{5})^2} \right )$ 

Karena, kita tahu bahwa, 
$2tan^{-1}x=tan^{-1}\left ( \frac{2x}{1-x^2} \right )$ 

$=tan\left (tan^{-1}\frac{\frac{2}{5}}{1-\frac{1}{25}} \right )$ 
$=tan\left (tan^{-1}\frac{5}{12} \right )$ $=\frac{5}{12}$ 

Contoh 2
Buktikan bahwa $4tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99}=\frac{\pi }{4}$ 

Jawab:
$=2(2tan^{-1}\frac{1}{5})-tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99}$

$=2\left (tan^{-1}\frac{2\times \frac{1}{5}}{1-(\frac{1}{5})^2} \right )-tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99} $ 
$=2\left (tan^{-1}\frac{\frac{2}{5}}{1-\frac{1}{25}} \right )-tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99} $ 
$=2tan^{-1}\frac{5}{12} -tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99} $ 
$=2tan^{-1}\left (\frac{2\times \frac{5}{12}}{1-(\frac{5}{12})^2} \right )-tan^{-1}\left ( \frac{\frac{1}{70}-\frac{1}{99}}{1+\frac{1}{70}\times \frac{1}{99}} \right )$ 
$=2tan^{-1}(\frac{120}{199} )-tan^{-1}\left ( \frac{29}{6931} \right )$ 
$=2tan^{-1}(\frac{120}{199} )-tan^{-1}\left ( \frac{1}{239} \right )$ 
$=tan^{-1}\left ( \frac{\frac{120}{199}-\frac{1}{239}}{1+\frac{120}{199}\times \frac{1}{239}} \right )$ 
$=tan^{-1}1$ 
$=tan^{-1}tan\frac{\pi }{4}=\frac{\pi }{4}$. TERBUKTI


Invers Fungsi Trigonometri


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