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

• Nilai Umum dan Pokok sin⁻¹ x
• Nilai Umum dan Pokok dari cos⁻¹ x
• Nilai Umum dan Pokok dari tan⁻¹ x
• Nilai Umum dan Pokok dari csc⁻¹ x
• Nilai Umum dan Pokok dari sec⁻¹ x
• Nilai Umum dan Pokok dari cot⁻¹ x
• Nilai Pokok Invers Fungsi Trigonometri  
• Nilai Umum Invers Fungsi Trigonometri  
• arcsin(x) + arccos(x) = 𝜋/2
• arctan(x) + arccot(x) = 𝜋/2
• $tan^{-1}x+tan^{-1}y=tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
• $tan^{-1}x-tan^{-1}y=tan^{-1}\left ( \frac{x-y}{1+xy} \right ) $
• $tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}\left ( \frac{x+y+z-xyz}{1-xy-yz-xz} \right ) $
• $cot^{-1}x+cot^{-1}y=cot^{-1}\left (\frac{xy-1}{y+x} \right ) $
• $cot^{-1}x-cot^{-1}y=cot^{-1}\left (\frac{xy+1}{y-x} \right ) $
• $sin^{-1}x+sin^{-1}y=sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$
• $sin^{-1}x-sin^{-1}y=sin^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2})$
• $cos^{-1}x+cos^{-1}y=cos^{-1}(xy-\sqrt{1-x^2} \sqrt{1-y^2})$
• $cos^{-1}x-cos^{-1}y=cos^{-1}(xy+\sqrt{1-x^2} \sqrt{1-y^2})$
• $2sin^{-1}x=sin^{-1}(2x\sqrt{1-x^2})$
• $2cos^{-1}x=cos^{-1}(2x^2-1)$
• $2tan^{-1}x=tan^{-1}\left ( \frac{2x}{1-x^2} \right )=sin^{-1}\left ( \frac{2x}{1+x^2} \right )=cos^{-1}\left ( \frac{1-x^2}{1+x^2} \right )$
• $3sin^{-1}x=sin^{-1}(3x-4x^3)$
• $3cos^{-1}x=cos^{-1}(4x^3-3x)$
• $3tan^{-1}x=tan^{-1}\left ( \frac{3x-x^3}{1-3x^2} \right )$
• Kumpulan Rumus Invers Fungsi Trigonometri  
• Nilai Utama Invers Fungsi Trigonometri
• Soal-soal & Pembahasan Invers Fungsi Trigonometri