Conversion from HSI to RGB

Consider three possible cases in terms of the hue angle $\angle H$:

• $0<\angle H < 120$ (p inside $\triangle P_rwP_g$)

  From $S=1-3b$, we get
  

  $$b=(1-S)/3$$
  
  
  Also we can get
  
  $$r=\frac{1}{3}[1+\frac{S cos\angle H}{cos(60-\angle H)}]$$
  
  
  and
  
  $$g=1-b-r$$
  
  
  Given $r,g,b$, we can get $R,G,B$ from $I$. As
  
  $$r=\frac{R}{R+G+B}=\frac{R}{3I}$$
  
  
  we have
  
  $$R=3Ir,\;\;\;G=3Ig,\;\;\;B=3Ib$$
  
  

• $120<\angle H < 240$ (p inside $\triangle P_gwP_b$)

  
  

  $$\angle H=\angle H-120$$
  
  
  
  
  $$r=(1-S)/3$$
  
  
  
  
  $$g=\frac{1}{3}[1+\frac{S cos\angle H}{cos(60-\angle H)}]$$
  
  
  
  
  $$b=1-g-r$$
  
  

• $240<\angle H < 360$ (p inside $\triangle P_bwP_r$)

  
  

  $$\angle H=\angle H-240$$
  
  
  
  
  $$g=(1-S)/3$$
  
  
  
  
  $$b=\frac{1}{3}[1+\frac{S cos\angle H}{cos(60-\angle H)}]$$
  
  
  
  
  $$r=1-g-b$$