a. Rumus untuk cos(a + b)
Perhatikan
Gambar
Dari gambar diperoleh OC =
OB = OA
= OD = r koordinat titik
A = (r,0),
B = (rcosa, rsina) ,
C = (rcos(a + b)
, rsin(a
+ b)),
dan
D = (rcos
b, rsin(- b )
Dengan menggunakan rumus jarak antara dua titik, diperoleh
d 2 AB = (AB)2 = (xB - x A)2 + (yB - y A)2
d 2 AB = (AB)2 = (xB - x A)2 + (yB - y A)2
AC2 =
DB2
2r 2 - 2r 2 cos(a + b) = 2r 2 + 2r 2 sina sin b - 2r 2 cosa cosb
- 2r 2 cos(a + b) = -2r 2(cosa cosb - sina sin b)
2r 2 - 2r 2 cos(a + b) = 2r 2 + 2r 2 sina sin b - 2r 2 cosa cosb
- 2r 2 cos(a + b) = -2r 2(cosa cosb - sina sin b)
Sehingga didapat
(AC)2 = [(r cos(a + b))- r]2 +((r sin(a + b))- 0)2
(AC)2 = r 2 cos 2(a + b)- 2r 2 cos(a + b)+ r 2 + r 2 sin 2(a + b)
(AC)2 = r 2[cos 2(a + b)+ sin 2(a + b)]- 2r 2 cos(a + b)+ r 2
(AC)2 = r 2 ×1+ r 2 - 2r 2 cos(a + b)
(AC)2 = 2r 2 - 2r 2 cos(a + b)
Dan
(DB)2 = (r cosa - r cos b)2 + (r sina + r sin b)2
(DB)2 = r 2 cos 2 a - 2r 2 cosa cos b + r 2 cos 2 b + r 2 sin 2 a + 2r2 sina sinb +r2 sin2 b
(DB)2 = r 2(cos 2 a + sin 2 a)+ r 2(cos 2 b + sin 2 b) + 2r2 sina sinb - 2r2 cosa cosb
(DB)2 = r 2 + r 2 + 2r 2 sina sin b - 2r 2 cosa cos b
(DB)2 = 2r 2 + 2r 2 sina sin b - 2r 2 cosa cos b
(AC)2 = [(r cos(a + b))- r]2 +((r sin(a + b))- 0)2
(AC)2 = r 2 cos 2(a + b)- 2r 2 cos(a + b)+ r 2 + r 2 sin 2(a + b)
(AC)2 = r 2[cos 2(a + b)+ sin 2(a + b)]- 2r 2 cos(a + b)+ r 2
(AC)2 = r 2 ×1+ r 2 - 2r 2 cos(a + b)
(AC)2 = 2r 2 - 2r 2 cos(a + b)
Dan
(DB)2 = (r cosa - r cos b)2 + (r sina + r sin b)2
(DB)2 = r 2 cos 2 a - 2r 2 cosa cos b + r 2 cos 2 b + r 2 sin 2 a + 2r2 sina sinb +r2 sin2 b
(DB)2 = r 2(cos 2 a + sin 2 a)+ r 2(cos 2 b + sin 2 b) + 2r2 sina sinb - 2r2 cosa cosb
(DB)2 = r 2 + r 2 + 2r 2 sina sin b - 2r 2 cosa cos b
(DB)2 = 2r 2 + 2r 2 sina sin b - 2r 2 cosa cos b
Karena segitiga OCA dan segitiga ODB kongruen, maka
AC2 = DB2
2r 2 - 2r 2 cos(a + b) = 2r 2 + 2r 2 sina sin b - 2r 2 cosa cosb
Sehingga didapat Rumus :
Bagus,materinya sangat lengkap.
BalasHapus