Mencari Trigonometri Sudut Rangkap

sin(α + β) = sin α cos β + cos α sin β

sin 2α = sin(α + α) = sin α cos α + cos α sin α = 2 sin α cos α

cos 2α = 1 – 2 sin² α

cos 2α = 2 cos² α - 1

cos 2 α = cos² α – sin² α

Mencari sin 3α

sin 3α = sin(α + 2α)     = sin α cos 2α + cos α sin 2α

= sin α . (1 – 2 sin² α) + cos α . 2 sin α cos α

= sin α – 2 sin³ α + 2 sin α . cos² α

= sin α – 2 sin³ α + 2 sin α . (1 – sin² α)

= sin α – 2 sin³ α + 2 sin α – 2 sin³ α

= 3 sin α – 4 sin³ α

Jadi sin 3α = 3 sin α – 4 sin³ α

Mencari cos 3α

cos(α + β) = cos α cos β - sin α sin β

cos 3α = cos(α + 2α)    = cos α cos 2α - sin α sin 2α

= cos α . (2 cos² α - 1) - sin α . 2 sin α cos α

= 2 cos³ α – cos α – 2 sin² α . cos α

= 2 cos³ α – cos α - (1 – cos² α). 2 cos α

= 2 cos³ α – cos α – 2 cos α + 2 cos³ α

= 4 cos³ α – 3 sin α

Jadi cos 3α = 4 cos³ α – 3 sin α

\[\begin{array}{l}
 \tan \left( {\alpha  + \beta } \right) = \frac{{\tan \alpha  + \tan \beta }}{{1 - \tan \alpha \tan \beta }} \\
 \tan 2\alpha  = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \\
 \tan 3\alpha  = \tan \left( {2\alpha  + \alpha } \right) \\
  \Leftrightarrow \tan 3\alpha  = \frac{{\tan 2\alpha  + \tan \alpha }}{{1 - \tan 2\alpha \tan \alpha }} \\
  \Leftrightarrow \tan 3\alpha  = \frac{{\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} + \tan \alpha }}{{1 - \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }}\tan \alpha }} \\
  \Leftrightarrow \tan 3\alpha  = \frac{{\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} + \frac{{\tan \alpha \left( {1 - {{\tan }^2}\alpha } \right)}}{{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 - {{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }} - \frac{{2{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}} \\
  \Leftrightarrow \tan 3\alpha  = \frac{{\frac{{2\tan \alpha  + \tan \alpha  - {{\tan }^3}\alpha }}{{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 - {{\tan }^2}\alpha  - 2{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}} \\
  \Leftrightarrow \tan 3\alpha  = \frac{{\frac{{3\tan \alpha  - {{\tan }^3}\alpha }}{{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 - 3{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}} \\
  \Leftrightarrow \tan 3\alpha  = \frac{{3\tan \alpha  - {{\tan }^3}\alpha }}{{1 - 3{{\tan }^2}\alpha }} \\
 \end{array}\]