Tolong ya semuannya pakai cara jgn asal jawab

16) lihat gambar
nilai cos (A + B) = cos a cos b - sin a sin b
cos (A + B) = 3/5 . 5/13 - 4/5 . 12/13
cos (A + B) = 15/ 65 - 48/ 65
cos (A + B) = -33/65

17) lihat gambar
[tex]tan \: (a - b) = \frac{tan \: a \: - \: tan \: b}{1 + \: tan \: a \: tan \: b} \\ tan \: (a - b) = \frac{ \frac{3}{4} - \frac{5}{12} }{1 + ( \frac{3}{4} )( \frac{5}{12}) } \\ tan \: (a - b) = \frac{ \frac{9 \: - \: 5}{12} }{1 + ( \frac{15}{48}) } \\ tan \: (a - b) = \frac{ \frac{4}{12} }{ \frac{63}{48} } \\ tan \: (a - b) = \frac{4}{12} \times \frac{48}{63} \\ tan \: (a - b) = \frac{16}{63} [/tex]

18) nilai tan 75°
tan 75 = tan (45 + 30)° = 2 + √3

[tex]tan \: (45 + 30) = \frac{tan \: 45 \: + \: tan \: 30}{1 - tan \: 45 \: tan \: 30 } \\ tan \: (45 \: + \: 30)= \frac{1 \: + \frac{1}{3} \sqrt{3} }{1 -( 1)( \frac{1}{3} \sqrt{3} )} \\ tan \: (45 + 30) = \frac{1 + \frac{1}{3} \sqrt{3} }{ 1 - \frac{1}{3} \sqrt{3} } \\ tan \: (45 + 30) = \frac{1 + \frac{1}{3} \sqrt{3} }{ 1 - \frac{1}{3} \sqrt{3} } \times \frac{1 + \frac{1}{3} \sqrt{3} }{1 + \frac{1}{3} \sqrt{3} } \\ tan \: (45 + 30) = \frac{1 + \frac{2}{3} \sqrt{3 } + \frac{1}{3} }{1 - \frac{1}{3} } \\ semua \: kali \: 3 \\ tan \: (45 + 30) = \frac{3+ 2 \sqrt{3} + 1 }{3 - 1} \\ tan \: (45 + 30) = \frac{4 + 2 \sqrt{3} }{2} \\ tan \: (45 + 30) = \: \frac{2(2 + \sqrt{3}) }{2} \\ tan \: (a \: + \: b) = (2 + \sqrt{3} )[/tex]

19) sin 225° = sin (180 + 45)°
sin 225° = sin 180 cos 45 + cos 180 sin 45
sin 225 = (0) (½√2) + (-1) (½√2)
sin 225 = -½√2

20) sin 105 - sin 15 =
[tex] = 2 \: cos \: \frac{1}{2} (105 + 15)sin \: \frac{1}{2} (105 - 15) \\ = 2 \: cos \: \frac{1}{2} (120)sin \: \frac{1}{2} (90) \\ = 2cos \: (60)sin \: (45) \\ = 2( \frac{1}{2} )( \frac{1}{2} \sqrt{2} ) \\ = \frac{1}{2} \sqrt{2} [/tex]