# Find i) cos(5  ) in terms of cos  ii) sin(5  ) in terms of sin  iii) tan(5  ) in terms of tan  cos(5  ) + isin(5  ) = (cos  + isin  ) 5 = cos.

## Presentation on theme: "Find i) cos(5  ) in terms of cos  ii) sin(5  ) in terms of sin  iii) tan(5  ) in terms of tan  cos(5  ) + isin(5  ) = (cos  + isin  ) 5 = cos."— Presentation transcript:

Find i) cos(5  ) in terms of cos  ii) sin(5  ) in terms of sin  iii) tan(5  ) in terms of tan  cos(5  ) + isin(5  ) = (cos  + isin  ) 5 = cos 5  + 5icos 4  sin  + 10i 2 cos 3  sin 2  + 10i 3 cos 2  sin 3  + 5i 4 cos  sin 4  + i 5 sin 5  = cos 5  – 10cos 3  sin 2  5cos  sin 4  + i(5cos 4  sin  – 10cos 2  sin 3  sin 5  cos5  = cos 5  – 10cos 3  sin 2  5cos  sin 4  Equating real parts sin5  = 5cos 4  sin  – 10cos 2  sin 3  sin 5  Equating imaginary parts Replace sin 2  by 1– cos 2  and sin 4  by (1–cos 2  2 in cos5  formula cos5  = cos 5  – 10cos 3  cos 2  5cos  cos 2    Expand and group terms Replace cos 2  by 1 – sin 2  and cos 4  by (1 – sin 2  ) 2 in sin5  formula sin5  = 5(1–sin 2   sin  – 10(1–sin 2  sin 3  sin 5  Expand and group terms Using de Moivres Theorem to find cos n  and sin n  Using De Moivres Theorem Using Pascals Triangle line 5 Separating Real and Imaginary Parts

tan5  = Divide top and bottom terms by cos 5  Cancel cos terms

Using de Moivres Theorem to express cos n  and sin n  in terms of cos  and sin  z = cos  +  isin  e i  z n = cos(n  + isin(n  e in  = (cos  + isin    = cos  isin  e –i  = cosn  – isin(n  e –in  As cos(–  = cos  and sin(–  ) = –sin 

Express cos 2  in terms of cos  = 2cos(2  ) + 2 as 4cos 2  = 2cos(2  ) + 2 cos 2  =  (cos(2  ) + 1)

1) Express sin 5  in terms of sin  2) Integrate the answer = 2isin5  – 5  2isin3  + 10  2isin  = 2isin5  – 10isin3  + 20isin  Pascals Triangle line 5 Grouping powers

32isin 5  = 2isin5  – 10isin3  + 20isin  sin 5  = Hence

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