arccos (x) + arccos (y) = arccos {(xy-√(1 – x^2) √(1 – y^2)}

Kita akan belajar bagaimana membuktikan sifat invers fungsi trigonometri$arccos\alpha +arccos\beta =arccos\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$

Misalkan, cos⁻¹ x = α dan cos⁻¹ y = 𝛽

Dari cos⁻¹ x = α diperoleh,

x = cos α

dan dari cos⁻¹ y = 𝛽 kita peroleh,

y = cos 𝛽

karena cos (α + 𝛽) = cosα cos 𝛽 – sin αsin 𝛽, maka$\Rightarrow cos(\alpha +\beta )=cos\alpha cos\beta - \sqrt{1-cos^2\alpha }.\sqrt{1-cos^2\beta }$ 

$\Rightarrow cos(\alpha +\beta )=xy - \sqrt{1-x^2 }.\sqrt{1-y^2 }$ 

$\Rightarrow \alpha +\beta =cos^{-1}\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$ $\Rightarrow cos^{-1}\alpha +cos^{-1}\beta =cos^{-1}\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$ 

Oleh karena itu $arccos\alpha +arccos\beta =arccos\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$

Catatan: Jika x > 0, y > 0 dan x² + y² > 1, maka cos⁻¹ x + sin⁻¹ y dapat berupa sudut lebih dari 𝜋/2 $cos^{-1}\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$ sedangkan adalah sudut antara – 𝜋/2 dan 𝜋/2.

karena itu  $cos^{-1}x +cos^{-1}y = \pi -cos^{-1}\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$

Contoh Soal 1

Jika , $cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{a}=\alpha$  Buktikan$\frac{x^2}{a^2}-\frac{2xy}{ab}cos\alpha +\frac{x^2}{b^2}=sin^2\alpha $

Jawab:

Kita punya $cos^{-1}x +cos^{-1}y = \pi -cos^{-1}\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )$

$\Rightarrow cos^{-1}\left [ \frac{x}{a}.\frac{y}{b}-\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}} \right ]=\alpha $ 

$\Rightarrow \left [ \frac{x}{a}.\frac{y}{b}-\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}} \right ]=cos\alpha $ 

$\Rightarrow \frac{xy}{ab}-cos\alpha =\sqrt{(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2})} $ 

$\Rightarrow \left (\frac{xy}{ab}-cos\alpha \right )^2 =(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2}) $ 

$\Rightarrow \frac{x^2y^2}{a^2b^2}-2\frac{xy}{ab}cos\alpha +cos^2\alpha =1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2y^2}{a^2b^2} $ 

$\Rightarrow \frac{x^2}{a^2}-2\frac{xy}{ab}cos\alpha +\frac{y^2}{b^2} =1-cos^2\alpha$ 

$\Rightarrow \frac{x^2}{a^2}-2\frac{xy}{ab}cos\alpha +\frac{y^2}{b^2} =sin^2\alpha$. Terbukti 

Contoh Soal 2

Jika cos⁻¹ x + cos⁻¹ y + cos⁻¹ z = , buktikan bahwa x² + y² + z² + 2xyz = 1.

Jawab:

cos⁻¹ x + cos⁻¹ y + cos⁻¹ z =

cos⁻¹ x + cos⁻¹ y = - cos⁻¹ z

cos⁻¹ x + cos⁻¹ y = cos⁻¹ (-z), [karena, cos⁻¹ (-θ) = 𝜋 - cos⁻¹ θ]$cos^{-1}\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )=cos^{-1}(-z)$

$\left (xy - \sqrt{1-x^2 }.\sqrt{1-y^2 } \right )=(-z)$

$xy +z= \sqrt{1-x^2 }.\sqrt{1-y^2 }$

sekarang kuadratkan kedua ruas

$\Rightarrow \left (xy +z \right )^2= (1-x^2)(1-y^2)$

$\Rightarrow x^2y^2+z^2+2xyz=1-x^2-y^2+x^2y^2$

$\Rightarrow x^2+y^2+z^2+2xyz=1 $. 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