Untuk
bilangan bulat positif, ekspresi dari bentuk
.
, Dan
dapat dinyatakan dalam hal
dan
hanya menggunakan rumus Euler dan teorema binomial . Untuk
.
Beberapa nilai pertama diberikan oleh
Formula terkait lainnya termasuk
di mana
adalah fungsi Floor .
Sebuah formula produk
diberikan oleh
Fungsi
juga dapat dinyatakan sebagai polinomial dalam
(Untuk
aneh) atau
kali polinomial di
sebagai
di mana
adalah polinomial Chebyshev jenis pertama dan
adalah polinomial Chebyshev jenis kedua . Beberapa kasus pertama adalah
Demikian pula,
dapat dinyatakan sebagai
kali polinomial di
sebagai
Beberapa kasus pertama adalah
Bromwich (1991) memberikan rumus
di mana
.
Untuk
, Rumus multiple-angle dapat diturunkan sebagai
Beberapa nilai-nilai yang pertama adalah
Formula terkait lainnya termasuk
Fungsi
juga dapat dinyatakan sebagai polinomial dalam
(Untuk
bahkan) atau
kali polinomial di
sebagai
Beberapa kasus pertama adalah
Demikian pula,
dapat dinyatakan sebagai polinomial dalam
sebagai
Beberapa kasus pertama adalah
Bromwich (1991) memberikan rumus
di mana
.
Beberapa formula multi-angle pertama
adalah
diberikan oleh Beyer (1987, hal. 139) sampai
.
Formula Multiple-sudut juga dapat ditulis dengan menggunakan hubungan recurrence
Mencari rumus sudut ganda sin (nĪ±), cos (nĪ±), tan (nĪ±)







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(1)
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(2)
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(10)
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(11)
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(12)
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Sebuah formula produk

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(13)
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(14)
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(15)
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(21)
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(22)
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(23)
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(24)
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Untuk

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(25)
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(26)
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(27)
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(28)
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(29)
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(30)
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(31)
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(32)
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(33)
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(34)
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(35)
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(36)
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(37)
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(39)
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(41)
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(42)
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(43)
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(44)
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(45)
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(46)
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(47)
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(48)
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Beberapa formula multi-angle pertama

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(49)
|
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(50)
|
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(51)
|

Formula Multiple-sudut juga dapat ditulis dengan menggunakan hubungan recurrence
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(52)
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(53)
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Mencari rumus sudut ganda sin (nĪ±), cos (nĪ±), tan (nĪ±)
1)
Sin 2Ī± = sin(Ī±+Ī±) =
sin Ī±.cos Ī± + cos Ī±. sin Ī± =
2.sin Ī±.cos Ī±
sin 2Ī± = 2sin Ī±.cos Ī± (terbukti)
sin 2Ī± = 2sin Ī±.cos Ī± (terbukti)
2)
Cos 2Ī± = cos(Ī±+Ī±) = cosĀ²Ī± - sinĀ²Ī± =
cosĀ²Ī± -(1- cosĀ²Ī±) = cosĀ²Ī± -1+cosĀ²Ī±
cos
2Ī± = 2cosĀ²Ī± - 1 (terbukti)
3)
Cos 2Ī± = cosĀ²Ī± - sinĀ²Ī± = (1 - sinĀ²Ī±) -
sinĀ²Ī± = 1- sinĀ²Ī± - sinĀ²Ī±
cos 2Ī± = 1-2sinĀ²Ī± (Terbukti)
cos 2Ī± = 1-2sinĀ²Ī± (Terbukti)
4)
Cos 2Ī± = cos(Ī±+Ī±) = cosĪ±.cosĪ± ā sinĪ±.sinĪ± =
cos 2Ī± = cosĀ²Ī± - sinĀ²Ī± (terbukti)
cos 2Ī± = cosĀ²Ī± - sinĀ²Ī± (terbukti)
5)
Tan 2Ī± = (tan Ī±+ tan Ī±)/(1-
tan Ī±. tan Ī±)
Tan 2Ī±= (2 tan Ī±)/(1-tanĀ²Ī±) (terbukti)
Tan 2Ī±= (2 tan Ī±)/(1-tanĀ²Ī±) (terbukti)
6)
Tan 2Ī± = sin 2Ī±/cos 2Ī± = (2sinĪ±.cosĪ±)/(cosĀ²Ī± -
sinĀ²Ī±) = {2[(sinĪ±.cosĪ±)/cosĀ²Ī±]}/{(cosĀ²Ī±/cosĀ²Ī±) - (sinĀ²Ī±/cosĀ²Ī±)} =
Tan 2Ī±= (2 tan Ī±)/(1-tanĀ²Ī±) (terbukti)
Tan 2Ī±= (2 tan Ī±)/(1-tanĀ²Ī±) (terbukti)
7)
Tan 2Ī± = {2tanĪ±}/(1-tanĀ²Ī±} =
{2tanĪ±(1/tanĪ±)}/{(1-tanĀ²Ī±)(1/tanĪ±)} = 2/{(1/tanĪ±) ā (tanĀ²Ī±/tanĪ±)} = 2/{cotĪ± ā
tanĪ±}
Tan 2Ī± =
2/{cotĪ± ā tanĪ±} (terbukti)
8)
Tan 2Ī± = 2/{cotĪ± ā tanĪ±} = (2 tan Ī±)/(1-tanĀ²Ī±)
= {2cotĪ±}/{cotĀ²Ī± ā 1}
9)
Cosec
2Ī± = 1/(sin 2Ī±) = 1/(2sin Ī±.cos Ī±) = 1/2 csc Ī±. sec Ī±
Cosec 2Ī± = 1/2 cosec Ī±. sec Ī± (Terbukti)
Cosec 2Ī± = 1/2 cosec Ī±. sec Ī± (Terbukti)
10)
cosec
2Ī± = cosec (Ī±+Ī±) = (cosecĪ±.cosecĪ±)/(cotan Ī±+cotan Ī±) =
(cosecĀ²Ī±)/(2cot Ī±)
cosec 2Ī± = (cosecĀ²Ī±)/(2cot Ī±) atau (cotĀ²Ī± + 1)/(2cot Ī±) (terbukti)
cosec 2Ī± = (cosecĀ²Ī±)/(2cot Ī±) atau (cotĀ²Ī± + 1)/(2cot Ī±) (terbukti)
11)
Cosec 2Ī± = 1/{2sinĪ±cosĪ±} = {cosecĀ²Ī±}/{2cotĪ±} =
secĀ²Ī±/2tanĪ± = {1+cotĀ²Ī±}/{2cotĪ±} = {1+tanĀ²Ī±}/{2tanĪ±}
12)
Cosec 2Ī± = 1/sin2Ī± = 1/(2sinĪ±cosĪ±) =
(cosĪ±)/(2sinĪ±cosĀ²Ī±) = {secĀ²Ī±}/{2(sinĪ±/cosĪ±)} = (secĀ²Ī±)/(2tanĪ±) =
(1+tanĀ²Ī±)/(2tanĪ±)
Cosec 2Ī± = (1+tanĀ²Ī±)/(2tanĪ±)
13)
Cosec
2Ī± = 1/sin2Ī± = (sinĀ²Ī± +
cosĀ²Ī±)/(2sinĪ±.cos Ī±) = 1/2
tan Ī± + cot Ī±
14)
Sec 2Ī±
= 1/(cos 2Ī±) = 1/(cosĀ²Ī± - sinĀ²Ī±) = 1/((1/secĀ²Ī±)-(1/cosecĀ²Ī±))
= 1/((cosecĀ²Ī±-secĀ²Ī±)/(secĀ²Ī±.cosecĀ²Ī±))
= (secĀ²Ī± . cosecĀ²Ī±)/(cosecĀ²Ī±-secĀ²Ī±)
Sec 2Ī± = (secĀ²Ī± . cosecĀ²Ī±)/(cosecĀ²Ī±-secĀ²Ī±) (Terbukti)
Sec 2Ī± = (secĀ²Ī± . cosecĀ²Ī±)/(cosecĀ²Ī±-secĀ²Ī±) (Terbukti)
15)
sec 2Ī± = sec(Ī±+Ī±) = (secĪ±.secĪ±)/(1-tanĪ±.tanĪ±) = (secĀ²Ī±)
/ (1-tanĀ²Ī±)
sec2Ī± = (secĀ²Ī±)/(1-tanĀ²Ī±) atau (1+tanĀ²Ī±) / (1-tanĀ²Ī±)(terbukti)
sec2Ī± = (secĀ²Ī±)/(1-tanĀ²Ī±) atau (1+tanĀ²Ī±) / (1-tanĀ²Ī±)(terbukti)
16)
sec 2Ī± = 1/{cosĀ²Ī±-sinĀ²Ī±} = 1/{2cosĀ²Ī±-1} = 1/{1-2sinĀ²Ī±}
17)
sec 2Ī± = {secĀ²Ī±}/{1-tanĀ²Ī±} = {cosecĀ²Ī±}/{cotĀ²Ī±-1} =
{1+tanĀ²Ī±}/{1-tanĀ²Ī±} = {1+cotĀ²Ī±}/{cotĀ²Ī±-1}
18)
sec 2Ī± = 1/cos 2Ī± = 1/(cosĀ²Ī±-sinĀ²Ī±) =
(cosĀ²Ī±+sinĀ²Ī±)/(cosĀ²Ī± - sinĀ²Ī±) = {(cosĀ²Ī±/cosĀ²Ī±)+(sinĀ²Ī±/cosĀ²Ī±)}/{(cosĀ²Ī±/cosĀ²Ī±)-(sinĀ²Ī±/cosĀ²Ī±)}
= {1+tanĀ²Ī±}/{1-tanĀ²Ī±}
sec 2Ī± = {1+tanĀ²Ī±}/{1-tanĀ²Ī±} terbukti
19)
cotan
2Ī± = 1/ tan 2Ī± = 1/ tan (Ī±+Ī±) = 1/ (2tan Ī±/(1-tanĀ²Ī±)) = 1-tanĀ²Ī± / 2 tanĪ±
cotan 2Ī± = 1-tanĀ²Ī± / 2 tanĪ±
cotan 2Ī± = 1-tanĀ²Ī± / 2 tanĪ±
* lanjutan cotan 2Ī±, = (1-(1/cotĀ²Ī±))/
(2.(1/cot Ī±))
= (cot Ī±.cotĪ± -1)/ (cotĪ± + cotĪ±) = (cotĀ²Ī± -1 /cotĀ²Ī±) x (cot Ī±/2)
= (cot Ī±.cotĪ± -1)/ (cotĪ± + cotĪ±) = (cotĀ²Ī± -1 /cotĀ²Ī±) x (cot Ī±/2)
cotan 2Ī± = (cotĀ²Ī±
-1)/(2 cot Ī±) atau (1-tanĀ²Ī±)/(2tan Ī±) atau (cot Ī±-tanĪ±)/2
(Terbukti)
20)
cotan
2Ī± = cot(Ī±+Ī±) = (cot Ī±.cot Ī± -1) / (cotĪ±+cotĪ±)
cotan 2Ī± = (cotĀ²Ī±
-1) / (2 cot Ī±) (terbukti)
21)
cotan 2Ī± = cos 2Ī±/sin 2Ī± = {cosĀ²Ī± -
sinĀ²Ī±}/{2sinĪ±cosĪ±} = (cosĪ±cosĪ±/2sinĪ±cosĪ±) ā (sinĪ±sinĪ±/2sinĪ±cosĪ±) = (cosĪ±/2sinĪ±)
ā (sinĪ±/2cosĪ±) = Ā½ (cotĪ± ā tanĪ±)
22)
cotan 2Ī± = cos 2Ī±/sin 2Ī± = (cosĀ²Ī± - sinĀ²Ī±)/ 2
(sinĪ±cosĪ±) = Ā½ {(cosĪ±/sinĪ±) ā (sinĪ±/cosĪ±)}
cotan 2Ī± = Ā½
(cotĪ± ā tanĪ±)
23)
cotan 2Ī± = {cotĪ± ā tanĪ±}/2 = {1-tanĀ²Ī±}/{2tanĪ±} =
{cotĀ²Ī±-1}/{2cotĪ±}
24)
Sin 3Ī± = sin (Ī± + 2Ī±)
= sin Ī± cos 2Ī± + sin 2Ī± cos Ī±
= sin Ī± (1 - 2 sinĀ² Ī±) + (2 sin Ī± cos Ī±) cos Ī±
= sin Ī± - 2 sinĀ³ Ī± + 2 sin Ī± cosĀ² Ī±
= sin Ī± - 2 sinĀ³ Ī± + 2 sin Ī± (1 - sinĀ² Ī±)
Sin 3Ī± = 3 sin Ī± - 4 sinĀ³ Ī± (terbukti)
= sin Ī± cos 2Ī± + sin 2Ī± cos Ī±
= sin Ī± (1 - 2 sinĀ² Ī±) + (2 sin Ī± cos Ī±) cos Ī±
= sin Ī± - 2 sinĀ³ Ī± + 2 sin Ī± cosĀ² Ī±
= sin Ī± - 2 sinĀ³ Ī± + 2 sin Ī± (1 - sinĀ² Ī±)
Sin 3Ī± = 3 sin Ī± - 4 sinĀ³ Ī± (terbukti)
25)
Sin 3Ī± = sin ( 2Ī± + Ī±)= sin 2Ī±.cos Ī± + cos2Ī±.sin Ī±
= (2sin Ī±.cos Ī±)cos Ī± +
(1-2.sinĀ²Ī±).sin Ī±
= 2.sin Ī±.cosĀ² Ī± + sin Ī± - 2. sinĀ³ Ī±
= 2.sin Ī±(1 ā sinĀ²Ī±) + sin Ī± ā2.sinĀ³Ī±
= 2.sin Ī± ā2.sin Ā³Ī± + sin Ī± ā2.sin Ā³Ī±
= 2.sin Ī± ā2.sin Ā³Ī± + sin Ī± ā2.sin Ā³Ī±
Sin 3Ī± =
3.sin Ī± ā4.sinĀ³Ī± (terbukti)
26)
Sin
3 Ī±= sin (2Ī±+Ī±)
= sin 2Ī± cos Ī± + cos 2Ī± sin Ī±
= 2 sin Ī± cos Ī± cos Ī± + (1-2 sinĀ² Ī±) sin Ī±
= 2 sin Ī± cosĀ²Ī± + sin Ī± - 2sinĀ³ Ī±
= 2 sin Ī±(1 - sinĀ²Ī±) + sin Ī± - 2sinĀ³ Ī±
= 2 sin Ī± - 2sinĀ³ Ī± + sin Ī± - 2sinĀ³ Ī±
Sin 3Ī± = 3 sin Ī± - 4 sinĀ³ Ī± (terbukti)
= sin 2Ī± cos Ī± + cos 2Ī± sin Ī±
= 2 sin Ī± cos Ī± cos Ī± + (1-2 sinĀ² Ī±) sin Ī±
= 2 sin Ī± cosĀ²Ī± + sin Ī± - 2sinĀ³ Ī±
= 2 sin Ī±(1 - sinĀ²Ī±) + sin Ī± - 2sinĀ³ Ī±
= 2 sin Ī± - 2sinĀ³ Ī± + sin Ī± - 2sinĀ³ Ī±
Sin 3Ī± = 3 sin Ī± - 4 sinĀ³ Ī± (terbukti)
27)
Sin (3Ī±)
= sin(Ī±+2Ī±) = sin(Ī±).cos(2Ī±) + cos(Ī±).sin(2Ī±)
= sin(Ī±)(cosĀ²(Ī±) ā sinĀ²(Ī±)) + cos(Ī±)(2sin(Ī±)cos(Ī±))
= sin(Ī±)(cosĀ²(Ī±) ā sinĀ²(Ī±)) + cos(Ī±)(2sin(Ī±)cos(Ī±))
= sin(Ī±)cosĀ²(Ī±) ā sinĀ³(Ī±) +
2sin(Ī±)cosĀ²(Ī±)
sin 3Ī± =
3sin(Ī±)cosĀ²(Ī±) ā sinĀ³(Ī±) (Terbukti)
28)
Sin
3Ī± = 3cosĀ²Ī±.sinĪ± ā sinĀ³Ī± = 3(1-sinĀ²Ī±)sinĪ± ā sinĀ³Ī± = 3sinĪ± ā 4sinĀ³Ī±
29)
Sin 3Ī±
= sin(Ī±+2Ī±) = sinĪ±.cos2Ī± + cosĪ±.sin2Ī± = sinĪ±(cosĀ²Ī± ā
sinĀ²Ī±) + cosĪ±(2sinĪ±.cosĪ±)
= cosĀ²Ī±.sinĪ± ā sinĀ³Ī± + 2cosĀ²Ī±.sinĪ± = 3cosĀ²Ī±.sinĪ±
- sinĀ³Ī± = sinĪ±(3cosĀ²Ī±-sinĀ²Ī±) = sinĪ±(3(1-sinĀ²Ī±)-sinĀ²Ī±) = sinĪ±(3-4sinĀ²Ī±) = 3sinĪ±
ā 4sinĀ³Ī±
Sin 3Ī±
= 3sinĪ± ā 4sinĀ³Ī±
30)
Cos
3Ī± = cosĀ³Ī± ā 3cosĪ±.sinĀ³Ī± = cosĀ³Ī± ā 3cosĪ±(1-cosĀ²Ī±) = 4cosĀ³Ī± ā 3cosĪ±
31)
Cos 3Ī± = cos (2Ī±+Ī±)
= Cos2Ī±.cosĪ± - sin2Ī±.SinĪ±
= (2cosĀ²Ī±-1)CosĪ± - 2sinĀ²Ī±.cosĪ±
= 2cosĀ³Ī± - CosĪ± - 2(1-cosĀ²Ī±)CosĪ±
= 2cosĀ³Ī± - CosĪ± - 2cosĪ± + 2cosĀ³Ī±
Cos 3Ī± = 4cosĀ³A - 3cosA (Terbukti)
= Cos2Ī±.cosĪ± - sin2Ī±.SinĪ±
= (2cosĀ²Ī±-1)CosĪ± - 2sinĀ²Ī±.cosĪ±
= 2cosĀ³Ī± - CosĪ± - 2(1-cosĀ²Ī±)CosĪ±
= 2cosĀ³Ī± - CosĪ± - 2cosĪ± + 2cosĀ³Ī±
Cos 3Ī± = 4cosĀ³A - 3cosA (Terbukti)
32)
Cos 3Ī±
= cos(Ī±+2Ī±) = cos(Ī±)cos(2Ī±) ā sin(Ī±)sin(2Ī±)
= cos(Ī±)(cosĀ²(Ī±) ā sinĀ²(Ī±)) ā sin(Ī±) x 2 sin(Ī±)cos(Ī±)
= cosĀ³(Ī±) ā sinĀ²(Ī±)cos(Ī±) ā 2 sinĀ²(Ī±)cos(Ī±)
= cos(Ī±)(cosĀ²(Ī±) ā sinĀ²(Ī±)) ā sin(Ī±) x 2 sin(Ī±)cos(Ī±)
= cosĀ³(Ī±) ā sinĀ²(Ī±)cos(Ī±) ā 2 sinĀ²(Ī±)cos(Ī±)
cos 3Ī± = cosĀ³(Ī±) ā 3
sinĀ²(Ī±)cos(Ī±) (terbukti)
33)
Cos 3Ī±
= cos(Ī±+2Ī±) = cosĪ±.cos2Ī± ā sinĪ±.sin2Ī± = cosĪ±(cosĀ²Ī± ā
sinĀ²Ī±) ā sinĪ±(2sinĪ±.cosĪ±)
= cosĀ³Ī± ā cosĪ±.sinĀ²Ī± ā 2cosĪ±.sinĀ²Ī± = cosĀ³Ī± ā 3cosĪ±.sinĀ²Ī± = cosĪ±(cosĀ²Ī± ā 3sinĀ²Ī±) = cosĪ±(cosĀ²Ī± ā 3(1-cosĀ²Ī±)) = cosĪ±(cosĀ²Ī± ā 3+3cosĀ²Ī±) = cos Ī± (4cosĀ²Ī±-3) = 4cosĀ³Ī± ā 3cosĪ±
= cosĀ³Ī± ā cosĪ±.sinĀ²Ī± ā 2cosĪ±.sinĀ²Ī± = cosĀ³Ī± ā 3cosĪ±.sinĀ²Ī± = cosĪ±(cosĀ²Ī± ā 3sinĀ²Ī±) = cosĪ±(cosĀ²Ī± ā 3(1-cosĀ²Ī±)) = cosĪ±(cosĀ²Ī± ā 3+3cosĀ²Ī±) = cos Ī± (4cosĀ²Ī±-3) = 4cosĀ³Ī± ā 3cosĪ±
Cos 3Ī± = 4cosĀ³Ī± ā 3cosĪ±
34)
Tan (3Ī±) = Tan (2Ī± + Ī±)
= (2tan Ī± / (1-tanĀ²Ī±) + Tan Ī±) / (1 - 2tanĀ²Ī± / (1-tanĀ²Ī±))
= (2tanĪ± + Tan Ī± (1-tanĀ²Ī±)) / ((1-tanĀ²Ī±) - 2tanĀ²Ī±)
= (2tanĪ± + Tan Ī± - tanĀ³Ī±) / (1 - 3tanĀ²Ī±)
Tan 3Ī± = (3tan Ī± - tanĀ³Ī±) / (1 - 3tanĀ²Ī±) (terbukti)
= (2tan Ī± / (1-tanĀ²Ī±) + Tan Ī±) / (1 - 2tanĀ²Ī± / (1-tanĀ²Ī±))
= (2tanĪ± + Tan Ī± (1-tanĀ²Ī±)) / ((1-tanĀ²Ī±) - 2tanĀ²Ī±)
= (2tanĪ± + Tan Ī± - tanĀ³Ī±) / (1 - 3tanĀ²Ī±)
Tan 3Ī± = (3tan Ī± - tanĀ³Ī±) / (1 - 3tanĀ²Ī±) (terbukti)
35)
Tan 3Ī± = sin 3Ī± / cos 3Ī±
= Sin (2Ī±+Ī±)/cos (2Ī±+Ī±)
= [Sin 2Ī±cos Ī± + cos2Ī±.sin Ī±] / [cos2Ī±cosĪ± - sin2Ī±sinĪ±]
= [2sin Ī±.cosĀ²Ī± + cos Ā²Ī±.sin Ī± - sinĀ³Ī±] / [cosĀ³Ī± - sinĀ²Ī±.cos Ī± - 2sinĀ²Ī±.cos Ī±]
= [3sin Ī±.cos Ā²Ī± - sin Ā³Ī±] / [cosĀ³Ī± - 3sinĀ²Ī±. cos Ī±]
Membagi pembilang dan penyebut dengan cos Ā³Ī± untuk mendapatkan:
= [3sin Ī±. cosĀ²Ī±/cosĀ³Ī± - sinĀ³Ī±/cosĀ³Ī±]/[cosĀ³Ī±/cosĀ³Ī± - 3sinĀ²Ī±.cosĪ±/cosĀ³Ī±]
= [3sin Ī± / Cos Ī± - tan Ā³Ī±] / [1 - 3sinĀ²Ī±/cosĀ²Ī±]
Tan 3Ī± = [3tan Ī± - tan Ā³Ī±] / [1 - 3tan Ā²Ī±] (terbukti)
= Sin (2Ī±+Ī±)/cos (2Ī±+Ī±)
= [Sin 2Ī±cos Ī± + cos2Ī±.sin Ī±] / [cos2Ī±cosĪ± - sin2Ī±sinĪ±]
= [2sin Ī±.cosĀ²Ī± + cos Ā²Ī±.sin Ī± - sinĀ³Ī±] / [cosĀ³Ī± - sinĀ²Ī±.cos Ī± - 2sinĀ²Ī±.cos Ī±]
= [3sin Ī±.cos Ā²Ī± - sin Ā³Ī±] / [cosĀ³Ī± - 3sinĀ²Ī±. cos Ī±]
Membagi pembilang dan penyebut dengan cos Ā³Ī± untuk mendapatkan:
= [3sin Ī±. cosĀ²Ī±/cosĀ³Ī± - sinĀ³Ī±/cosĀ³Ī±]/[cosĀ³Ī±/cosĀ³Ī± - 3sinĀ²Ī±.cosĪ±/cosĀ³Ī±]
= [3sin Ī± / Cos Ī± - tan Ā³Ī±] / [1 - 3sinĀ²Ī±/cosĀ²Ī±]
Tan 3Ī± = [3tan Ī± - tan Ā³Ī±] / [1 - 3tan Ā²Ī±] (terbukti)
36)
Tan
3Ī± = tan(2Ī±+Ī±) = (tan2Ī±+tanĪ±)/(1- tan2Ī±.tanĪ±)
= {[(2tanĪ±)/(1-tanĀ² Ī±)]+ tan Ī±}/{1-(2tan Ī±/1-tanĀ² Ī±)(tan Ī±)}
= {[(2tan Ī±)/(1-tanĀ² Ī±)]+{[(1-tanĀ² Ī±)tan Ī±]/(1-tanĀ² Ī±)}/{[(1-tanĀ² Ī±)/ (1-tanĀ² Ī±)]-[2tanĀ² Ī±/(1-tanĀ² Ī±)]}
= {[(2tanĪ± + tanĪ± - tanĀ³Ī±)/(1-tanĀ²Ī±)]/[(1-tanĀ²Ī± - 2tanĀ²Ī±)/(1 - tanĀ²Ī±)]}
= {2tanĪ± + tanĪ± - tanĀ³Ī±}/{1 - tanĀ²Ī± - 2tanĀ²Ī±}
Tan 3Ī± = {3tan Ī± ā tanĀ³Ī±}/{1-3tanĀ²Ī±} (Terbukti)
= {[(2tanĪ±)/(1-tanĀ² Ī±)]+ tan Ī±}/{1-(2tan Ī±/1-tanĀ² Ī±)(tan Ī±)}
= {[(2tan Ī±)/(1-tanĀ² Ī±)]+{[(1-tanĀ² Ī±)tan Ī±]/(1-tanĀ² Ī±)}/{[(1-tanĀ² Ī±)/ (1-tanĀ² Ī±)]-[2tanĀ² Ī±/(1-tanĀ² Ī±)]}
= {[(2tanĪ± + tanĪ± - tanĀ³Ī±)/(1-tanĀ²Ī±)]/[(1-tanĀ²Ī± - 2tanĀ²Ī±)/(1 - tanĀ²Ī±)]}
= {2tanĪ± + tanĪ± - tanĀ³Ī±}/{1 - tanĀ²Ī± - 2tanĀ²Ī±}
Tan 3Ī± = {3tan Ī± ā tanĀ³Ī±}/{1-3tanĀ²Ī±} (Terbukti)
37)
Tan
3Ī± = sin 3Ī±/cos 3Ī± = {3cosĀ²Ī±.sinĪ± ā sinĀ³Ī±}/{cosĀ³Ī± ā 3cosĪ±sinĀ³Ī±} = {sinĪ±(3cosĀ²Ī±
-sinĀ²Ī±)}/{cosĪ±(cosĀ²Ī± -3sinĀ²Ī±)} = {sinĪ±(3-tanĀ²Ī±)}/{cosĪ±(1-3tanĀ²Ī±)} =
{(sinĪ±/cosĪ±)(3-tanĀ²Ī±)}/{1-3tanĀ²Ī±} = {tanĪ±(3-tanĀ²Ī±)}/(1-3tanĀ²Ī±)
Tan 3Ī± = {3tanĪ± ā tanĀ³Ī±}/{1-3tanĀ²Ī±}
38)
Tan
3Ī± = tan(2Ī±+Ī±) = (tan2Ī± + tanĪ±)/(1- tan2Ī±.tanĪ±) = {([2tanĪ±/1-tanĀ²Ī±]+
tan Ī±)(1-tanĀ²Ī±}/[{1-(2tan Ī±/1-tanĀ²Ī±)(tanĪ±)}(1-tanĀ²Ī±)] = = {2tanĪ± + tanĪ± -
tanĀ³Ī±}/{1 - tanĀ²Ī± - 2tanĀ²Ī±} = {3tanĪ± ā tanĀ³Ī±}/{1-3tanĀ²Ī±}
39)
cosec
3Ī± = 1/(sin 3Ī±) = 1/(3sin Ī± - 4sinĀ³Ī±)
atau (cosecĀ³Ī±)/(3cotĀ²Ī± - 1)
40)
sec 3Ī±
= 1 /cos (3Ī±) = 1 /(4cosĀ³Ī± - 3cosĪ±) atau (secĀ³Ī±)/(1 - 3tanĀ²Ī±)
41)
Cot
3Ī± = cos 3Ī±/sin 3Ī± = {cosĀ³Ī± ā 3cosĪ±sinĀ³Ī±}/{3cosĀ²Ī±.sinĪ± ā sinĀ³Ī±} = {cotĀ³Ī± ā 3cot Ī±}/{3cotĀ²Ī± ā 1}
42)
Cotan
3Ī±, ingat Tan 3Ī± = {tanĪ± + tan 2Ī±}/{1 ā tanĪ± .tan 2Ī±} = {(tan Ī±+ 2tanĪ±)/(1 ā
tanĀ²Ī±)}/{(1 ā tanĪ±.2tanĪ±)/(1 ā tanĀ²Ī±)} = {tan Ī±(1 ā tanĀ²Ī±)+2tanĪ±}/{1 ā tanĀ²Ī± ā
2tanĀ²Ī±} = {3tan Ī± ā tanĀ³Ī±}/{1 ā 3tanĀ²Ī±} = {3 cotĀ²Ī± ā 1}/{cotĀ³Ī± ā 3cot Ī±}
Sehingga
Cot 3Ī± = 1/tan3Ī± = {cotĀ³Ī± ā 3cot Ī±}/{3 cotĀ²Ī± ā 1} Terbukti
43)
Cotan
3Ī± = cos(3 Ī±)/sin(3 Ī±)
= [4cosĀ³(Ī±) - 3cos(Ī±)]/[3sin(Ī±) -
4sinĀ³(Ī±)]
Membagi kedua pembilang dan penyebut dengan sinĀ³(Ī±)
= {4(cosĀ³(Ī±)/sinĀ³(Ī±)) - 3(cos(Ī±)/sinĀ³(Ī±))}/{3(sin(Ī±)/sinĀ³(Ī±))-4(sinĀ³(Ī±)/sinĀ³(Ī±))}
= {4cotĀ³(Ī±) - 3(cos(Ī±)/sin(Ī±))(1/sinĀ²(Ī±))}/{3(1/sinĀ²(Ī±)) - 4}
= [4cotĀ³(Ī±) - 3cot(Ī±)cscĀ²(Ī±)]/[3cscĀ²(Ī±) - 4]
Ingat: cscĀ²(Ī±) - cotĀ²(Ī±) = 1, cot(Ī±) = cos(Ī±)/sin(Ī±),
= [4cotĀ³(Ī±) - 3cot(Ī±){1 + (cotĀ²(Ī±))}]/[3{1 + (cotĀ²(Ī±))} - 4]
Membagi kedua pembilang dan penyebut dengan sinĀ³(Ī±)
= {4(cosĀ³(Ī±)/sinĀ³(Ī±)) - 3(cos(Ī±)/sinĀ³(Ī±))}/{3(sin(Ī±)/sinĀ³(Ī±))-4(sinĀ³(Ī±)/sinĀ³(Ī±))}
= {4cotĀ³(Ī±) - 3(cos(Ī±)/sin(Ī±))(1/sinĀ²(Ī±))}/{3(1/sinĀ²(Ī±)) - 4}
= [4cotĀ³(Ī±) - 3cot(Ī±)cscĀ²(Ī±)]/[3cscĀ²(Ī±) - 4]
Ingat: cscĀ²(Ī±) - cotĀ²(Ī±) = 1, cot(Ī±) = cos(Ī±)/sin(Ī±),
= [4cotĀ³(Ī±) - 3cot(Ī±){1 + (cotĀ²(Ī±))}]/[3{1 + (cotĀ²(Ī±))} - 4]
cotan
3Ī± = [cotĀ³(Ī±) - 3cot(Ī±)]/[3cotĀ²(Ī±) - 1]
cotan 3Ī± = (3cotĪ± -
cotĀ³Ī±)/ (1 - 3cotĀ²Ī±) atau (1 - 3tanĀ²Ī±)/ (3tan Ī± - tanĀ³Ī±)
(terbukti)
44)
Cot
(Ī±+Ī²+Ļ) = {Re (i+cot Ī±)(i+cot Ī²)(i+cot c)}/{Im (i+cot Ī±)(i+cot Ī²)(i+cot c)}
=
{cotĪ±.cotĪ².cotĻ ā
cotĪ± ā cotĪ² ā cotĻ} / {cotĪ±.cotĪ² +
cotĪ±.cotĻ +
cotĪ².cot Ļ
ā 1}
Cot 3Ī± = {cotĀ³Ī± ā 3cot Ī±}/ {3cotĀ²Ī± ā 1} Terbukti
45)
Cot 3Ī± = cot (2Ī±+Ī±)
Dalam tan Ī± = 1/(tan (2Ī±+Ī±) = {1/(tan2Ī±+tan
Ī±)}/{1-tan2Ī±.tanĪ±} = (1-tan2Ī±.tanĪ±)/(tan 2Ī±+tan Ī±)
Dalam cot Ī± = {1-(1/cot 2Ī±)(1/cot Ī±)}/{(1/cot 2Ī±) + (1/cot
Ī±)} = {(cot 2Ī±.cot Ī± ā 1)/(cot 2Ī±.cot Ī±)}/{(cot Ī±+cot 2Ī±)/(cot 2Ī±.cot Ī±)} =
(cot 2Ī±.cot Ī± ā 1)/(cot 2Ī± + cot Ī±)
Cot 3É = (cot 2Ī±.cot Ī± ā 1)/(cot
2Ī± + cot Ī±)
46)
Cot 3Ī± = cos 3Ī±/sin 3Ī±
= {4cosĀ³Ī± ā 3cosĪ±}/{3sinĪ± ā 4sinĀ³Ī±} = {[4cosĀ³Ī± ā
3cosĪ±]/cosĀ³Ī±}x{1/[3(sinĪ±/cosĀ³Ī±)-4(sinĀ³Ī±/cosĀ³Ī±)]}= {4-(3/cosĀ²Ī±)}/{1/[3(sinĪ±/cosĪ±)(1/cosĀ²Ī±)-4tanĀ³Ī±]}
= {4-3(1+tanĀ²Ī±)}/{3tan Ī±(1+tanĀ²Ī±)-4tanĀ³Ī±} = {4-3-tanĀ²Ī±}/{3tanĪ±+3tanĀ³Ī± ā 4tanĀ³Ī±}
= {1-(1/cotĀ²Ī±)}/{(3/cotĪ±) ā (1/cotĀ³Ī±)} = (cotĀ²Ī±-1/cotĀ²Ī±)x(cotĀ³Ī±/3cotĀ²Ī±-1)
Cot 3Ī± = (cotĀ³Ī± ā
3cotĪ±)/(3cotĀ²Ī±-1)
47)
Sin 4Ī± = sin 2(2Ī±) = 2sin2Ī±.cos2Ī± =
2(2sinĪ±.cosĪ±)(cosĀ²Ī± - sinĀ²Ī±)
sin 4Ī± = 4sinĪ±.cosĀ³Ī± - 4sinĀ³Ī±.cosĪ± (terbukti)
sin 4Ī± = 4sinĪ±.cosĀ³Ī± - 4sinĀ³Ī±.cosĪ± (terbukti)
48)
sin 4a
= sin 2(2Ī±) = 2sin2Ī±.cosĪ± = 2(2sinĪ±.cosĪ±)(1-2sinĀ²Ī±)
sin 4Ī± = 4 sin Ī± cos Ī± - 8 sinĀ³ Ī± cos Ī± (terbukti)
sin 4Ī± = 4 sin Ī± cos Ī± - 8 sinĀ³ Ī± cos Ī± (terbukti)
49)
Sin 4Ī± = 4cosĀ³Ī±sinĪ± ā 4 cosĪ±sinĀ³Ī± =
4cosĪ±(1-sinĀ²Ī±)sinĪ± ā 4cosĪ±sinĀ³Ī± = 4cosĪ±(sinĪ±
ā 2sinĀ³Ī±)
50)
Sin 4Ī± = sin 2(2Ī±) = 2sin2Ī±.cos2Ī±
= 2(2sinĪ±.cosĪ±)(cosĀ²Ī± -
sinĀ²Ī±) = 4cosĪ±.sinĪ±(cosĀ²Ī± - sinĀ²Ī±) = 4cosĀ³Ī±.sinĪ± ā 4cosĪ±.sinĀ³Ī± = 4sinĪ±.cosĪ±(cosĀ²Ī±-sinĀ²Ī±) =
4sinĪ±.cosĪ±{(1-sinĀ²Ī±)-sinĀ²Ī±} = 4sinĪ±.cosĪ±(1-2sinĀ²Ī±) = 4sinĪ±.cosĪ± ā 8sinĀ³Ī±.cosĪ±
Sin 4Ī± = 4sinĪ±.cosĪ± ā 8sinĀ³Ī±.cosĪ±
51)
Cos 4Ī± = cosā“Ī± ā 6cosĀ²Ī±sinĀ²Ī± + sinā“Ī± = cosā“Ī± ā 6cosĀ²Ī±(1 ā
cosĀ²Ī±)+(1 ā cosĀ²Ī±)Ā² =
Cos
4Ī± = 8cosā“Ī±
ā 8cosĀ²Ī± + 1
52)
Cos 4Ī± = cos 2(2Ī±) = 2cosĀ²2Ī± ā 1 = 2(2cosĀ²Ī± -
1)Ā² - 1 = 2(4cosā“Ī± -
2cosĀ²Ī± + 1) - 1
= 8 cosā“Ī± - 8 cosĀ² Ī± + 2 - 1
cos 4Ī± = 8 cosā“Ī± ā 8 cosĀ² Ī± + 1 (Terbukti)
= 8 cosā“Ī± - 8 cosĀ² Ī± + 2 - 1
cos 4Ī± = 8 cosā“Ī± ā 8 cosĀ² Ī± + 1 (Terbukti)
53)
cos 4Ī± = 1-2sinĀ²2Ī± = 1-2(2sinĪ±cosĪ±)Ā² = 1-8sinĀ²Ī±cosĪ±
= 1-8sinĀ²Ī±(1-sinĀ²Ī±)
cos 4Ī± = 1-8sinĀ²Ī± + 8sinā“Ī± (terbukti)
54)
cos 4Ī±
= cos (2Ī±+2Ī±) = cos2Ī±.cos2Ī± ā sin2Ī±.sin2Ī± = cosĀ²2Ī± ā sinĀ²2Ī±
= (2cosĀ²Ī± ā 1)Ā² - (2sinĪ±cosĪ±)Ā²
= 4cosā“Ī± ā 4cosĀ²Ī± +1 ā 4sinĀ²Ī±cosĀ²Ī±
= 4cosā“Ī± ā 4cosĀ²Ī± +1 ā 4(1 ā cosĀ²Ī±)cosĀ²Ī± =
4cosā“Ī± ā 4cosĀ²Ī± +1 ā 4(cosĀ²Ī± - cosā“Ī±) = 4cosā“Ī± ā 4cosĀ²Ī± +1 ā 4cosĀ²Ī± +4cosā“Ī±
Cos 4Ī± =
8cosā“Ī± ā 8cosĀ²Ī± +1 (terbukti)
55)
cos 4Ī± = cos (2Ī±+2Ī±) = cos 2Ī± cos 2Ī± ā sin2Ī± sin2Ī±
= cosĀ²2Ī± ā sinĀ²2Ī± = (cosĀ²Ī± - sinĀ²Ī±)Ā² - (2sinĪ±.cosĪ±)Ā² = cosā“Ī± - 2cosĀ²Ī±.sinĀ²Ī± + sinā“Ī± - 4cosĀ²Ī±.sinĀ²Ī± = cosā“Ī± ā 6cosĀ²Ī±.sinĀ²Ī± + sinā“Ī± = cosā“Ī± - 6cosĀ²Ī±.sinĀ²Ī± +
(sinĀ²Ī±)Ā² = cosā“Ī± ā 6cosĀ²Ī±(1-cosĀ²Ī±) + (1-cosĀ²Ī±)Ā² =
cosā“Ī± ā 6cosĀ²Ī± + 6cosā“Ī± + 1 ā 2cosĀ²Ī± + cosā“Ī± = 8cosā“Ī± ā 8cosĀ²Ī± + 1
cos 4Ī± = 8cosā“Ī± ā 8cosĀ²Ī± + 1
56)
cos 4Ī±
= cos 2(2Ī±) = cosĀ²2Ī± - sinĀ²2Ī± = (cosĀ²Ī± -
sinĀ²Ī±)Ā² - (2sinĪ±.cosĪ±)Ā²
= cosā“Ī± - 2sinĀ²Ī±.cosĀ²Ī± + sinā“Ī± - 4sinĀ²Ī±.cosĀ²Ī±
= cosā“Ī± - 2sinĀ²Ī±.cosĀ²Ī± + sinā“Ī± - 4sinĀ²Ī±.cosĀ²Ī±
57) 57) Kita tahu tan 2Ī± = 2tan Ī± / (1-tanĀ²Ī±)
Dengan ini kita dapat menulis,
Tan 4Ī± = tan2 (2Ī±)
= 2 tan2A / (1- tanĀ² 2Ī±) ............ (1)
tan2Ī± = 2tan Ī± / (1- tanĀ² Ī±) ............... (2)
Padukan (2) dalam (1)
2 [2tan Ī± / (1- tanĀ² Ī±)]
= -------------------------------------...
1- [2 Tan Ī± / (1- tanĀ²Ī±)]Ā²
4 Tan Ī± / (1- tanĀ²Ī±)
= ----------------------------------------...
[(1- tanĀ²Ī±)Ā² - 4tanĀ²Ī±]/(1-tanĀ²Ī±)Ā²
[Salib multiply di penyebut, sekarang salah satu 1- tanĀ²Ī± akan dibatalkan]
4tan Ī±
= ----------------------------------------...
[1 + tanā“Ī± - 2tanĀ²Ī± 4tanĀ²Ī±] / (1-tanĀ²Ī±)
Penyebut dari penyebut adalah pembilang
= 4tan Ī± (1- tanĀ²Ī±) / [1 - 6tanĀ²Ī± + tanā“Ī±]
Tan 4Ī± = (4 Tan Ī± - 4 tanĀ³Ī±) / [1- 6tanĀ²Ī± + tanā“Ī±] (terbukti)
Dengan ini kita dapat menulis,
Tan 4Ī± = tan2 (2Ī±)
= 2 tan2A / (1- tanĀ² 2Ī±) ............ (1)
tan2Ī± = 2tan Ī± / (1- tanĀ² Ī±) ............... (2)
Padukan (2) dalam (1)
2 [2tan Ī± / (1- tanĀ² Ī±)]
= -------------------------------------...
1- [2 Tan Ī± / (1- tanĀ²Ī±)]Ā²
4 Tan Ī± / (1- tanĀ²Ī±)
= ----------------------------------------...
[(1- tanĀ²Ī±)Ā² - 4tanĀ²Ī±]/(1-tanĀ²Ī±)Ā²
[Salib multiply di penyebut, sekarang salah satu 1- tanĀ²Ī± akan dibatalkan]
4tan Ī±
= ----------------------------------------...
[1 + tanā“Ī± - 2tanĀ²Ī± 4tanĀ²Ī±] / (1-tanĀ²Ī±)
Penyebut dari penyebut adalah pembilang
= 4tan Ī± (1- tanĀ²Ī±) / [1 - 6tanĀ²Ī± + tanā“Ī±]
Tan 4Ī± = (4 Tan Ī± - 4 tanĀ³Ī±) / [1- 6tanĀ²Ī± + tanā“Ī±] (terbukti)
58)
Tan
4Ī± = sin 4Ī±/cos 4Ī± = {4cosĀ³Ī±sinĪ± ā 4 cosĪ±sinĀ³Ī±}/{cosā“Ī± ā 6cosĀ²Ī±sinĀ²Ī± + sinā“Ī±} = {4(cosĀ³Ī±sinĪ±/cosā“Ī±)-4(cosĪ±sinĀ³Ī±/cosā“Ī±)}/{(cosā“Ī±/cosā“Ī±) ā 6(cosĀ²Ī±sinĀ²Ī±/cosā“Ī±)+(sinā“Ī±/cosā“Ī±)} =
Tan 4É = {4tanĪ± ā 4 tanĀ³Ī±}/{1-6tanĀ²Ī± +
tanā“Ī±} = {4tanĪ±(1-tanĀ²Ī±)}/{1-6tanĀ²Ī± + tanā“Ī±}
59)
Tan4Ī±
= ingat tan 2Ī± = {2tanĪ±}/{1-tanĀ²Ī±} maka,
Tan4Ī± = tan 2(2Ī±) =
{2tan2Ī±}/{1-tanĀ²2Ī±} = {2(2tan Ī±/1-tanĀ²Ī±)}/[1-{2tan Ī±/(1-tanĀ²Ī±)}Ā²] = (4tan
Ī±/(1-tanĀ²Ī±)) / [1-{4tanĀ²Ī±/(1-tanĀ²Ī±)Ā²}] = {4tan Ī±/(1-tanĀ²Ī±)} / [{(1-tanĀ²Ī±)Ā² -
4tanĀ²Ī±}/(1-tanĀ²Ī±)Ā²] = {4tan Ī± (1-tanĀ²Ī±)}/{(1-tanĀ²Ī±)Ā² - 4 tanĀ²Ī±} = {4tan Ī±
(1-tanĀ²Ī±)}/{1+tanā“Ī± ā 2tanĀ²Ī± ā 4tanĀ²Ī±}
Tan 4É = {4tan
Ī± ā 4tanĀ³Ī±}/{1 ā 6tanĀ²Ī± +tanā“Ī±}
60)
Cot 4Ī± = ingat teorema de moivre dimana n integer
(cos Ī±+isinĪ±)āæ = cos (nĪ±) + isin(nĪ±)
ān = 4
Maka (cos Ī±+isinĪ±)ā“ = cos 4Ī± + isin 4Ī±
= cosā“Ī± + 4icosĀ³Ī±sinĪ± - 6cosĀ²Ī±sinĀ²Ī± ā 4i cosĪ±sinĀ³Ī± + sinā“Ī± = {cosā“Ī± ā
6cosĀ²Ī±sinĀ²Ī± + sinā“Ī±} + i{4cosĀ³Ī±sinĪ± ā 4 cosĪ±sinĀ³- }
Cotan 4Ī± = cos 4Ī±/sin 4Ī± = {cosā“Ī± ā
6cosĀ²Ī±sinĀ²Ī± + sinā“Ī±}/{4cosĀ³Ī±sinĪ± ā 4 cosĪ±sinĀ³Ī±} = {(cosā“Ī±/sinā“Ī±) ā (6cosĀ²Ī±sinĀ²Ī±/
sinā“Ī±)+(sinā“Ī±/sinā“Ī±)}/{4(cosĀ³Ī±sinĪ±/sinā“Ī±) ā 4(cosĪ±sinĀ³Ī±/sinā“Ī±)}
cotan
4Ī± = {cotā“Ī± ā 6cotĀ²Ī± + 1}/{4cotĀ³Ī± ā 4cotĪ±}
61)
Sin 5Ī± = sin5Ī± + sinĪ± -
sinĪ±
= 2 sin3Ī± cos2Ī± - sinĪ±
= 2 (3sinĪ± - 4sinĀ³ Ī±) (1 - 2sinĀ² Ī±) - sinĪ±
= 2 (8sināµ Ī± - 10sinĀ³ Ī± + 3sinĪ±) - sinĪ±
Sin 5Ī± = 16sināµ Ī± - 20sinĀ³ Ī± + 5sinĪ± (Terbukti)
= 2 sin3Ī± cos2Ī± - sinĪ±
= 2 (3sinĪ± - 4sinĀ³ Ī±) (1 - 2sinĀ² Ī±) - sinĪ±
= 2 (8sināµ Ī± - 10sinĀ³ Ī± + 3sinĪ±) - sinĪ±
Sin 5Ī± = 16sināµ Ī± - 20sinĀ³ Ī± + 5sinĪ± (Terbukti)
62)
Sin 5Ī± = sin(4Ī± + Ī±) = sin4Ī± cosĪ± +
cos4Ī± sinĪ±
= 2 sin2Ī± cos2Ī± cosĪ± + (1 - 2sinĀ² 2Ī±) sinĪ±
= 4sinĪ± cosĪ± cos2Ī± cosĪ± + sinĪ± - 2sinĀ² Ī± sinĪ±
= 4 sinĪ± cosĀ² Ī± (1 - 2sinĀ² Ī±) + sinĪ± - 2 (2sinĪ± cosĪ±)Ā² sinĪ±
= 4 sinĪ± (1 - sinĀ² Ī±) (1 - 2sinĀ² Ī±) + sinĪ± - 2 (4sinĀ² Ī± cosĀ² Ī±) sinĪ±
= 4 sinĪ± (1 - 3 sinĀ² Ī± + 2sinā“ Ī±) + sinĪ± - 2 [4 sinĀ³ Ī± * (1 - sinĀ² Ī±)]
= 4 sinĪ± - 12sinĀ³ Ī± + 8sināµ Ī± + sinĪ± - 8sinĀ³ Ī± + 8 sināµ Ī±
Sin 5Ī± = 16 sināµ Ī± - 20 sinĀ³ Ī± + 5 sinĪ±. (Terbukti)
= 2 sin2Ī± cos2Ī± cosĪ± + (1 - 2sinĀ² 2Ī±) sinĪ±
= 4sinĪ± cosĪ± cos2Ī± cosĪ± + sinĪ± - 2sinĀ² Ī± sinĪ±
= 4 sinĪ± cosĀ² Ī± (1 - 2sinĀ² Ī±) + sinĪ± - 2 (2sinĪ± cosĪ±)Ā² sinĪ±
= 4 sinĪ± (1 - sinĀ² Ī±) (1 - 2sinĀ² Ī±) + sinĪ± - 2 (4sinĀ² Ī± cosĀ² Ī±) sinĪ±
= 4 sinĪ± (1 - 3 sinĀ² Ī± + 2sinā“ Ī±) + sinĪ± - 2 [4 sinĀ³ Ī± * (1 - sinĀ² Ī±)]
= 4 sinĪ± - 12sinĀ³ Ī± + 8sināµ Ī± + sinĪ± - 8sinĀ³ Ī± + 8 sināµ Ī±
Sin 5Ī± = 16 sināµ Ī± - 20 sinĀ³ Ī± + 5 sinĪ±. (Terbukti)
63)
Cos 5Ī± = cos 5Ī± + cosĪ± -
cosĪ±
= 2 cos3Ī± cos2Ī± - cosĪ±
= 2 (4cosĀ³ Ī± - 3cosĪ±) (2cosĀ² Ī± - 1) - cos Ī±
= 2 (8cosāµĪ± - 10cosĀ³ Ī± + 3cos Ī±) - cos Ī±
Cos 5Ī± = 16cosāµ Ī± - 20cosĀ³ Ī± + 5cosĪ± (terbukti).
= 2 cos3Ī± cos2Ī± - cosĪ±
= 2 (4cosĀ³ Ī± - 3cosĪ±) (2cosĀ² Ī± - 1) - cos Ī±
= 2 (8cosāµĪ± - 10cosĀ³ Ī± + 3cos Ī±) - cos Ī±
Cos 5Ī± = 16cosāµ Ī± - 20cosĀ³ Ī± + 5cosĪ± (terbukti).
64)
Cos 5Ī± = cos(2Ī±
+3Ī±) = cos
2Ī±.cos 3Ī± ā sin 2Ī±.sin 3Ī± = (2cosĀ²Ī± ā 1) (4cosĀ³Ī± -3cosĪ±) ā 2sinĪ±.cosĪ± (3sinĪ± ā
4sinĀ³Ī±) = 8cosāµĪ± - 10cosĀ³Ī± + 3cosĪ± ā 6cosĪ±sinĀ²Ī± +8cosĪ±sinā“Ī± = 8cosāµĪ±
- 10cosĀ³Ī± + 3cosĪ± ā 6cosĪ±(1-cosĀ²Ī±)
+ 8cos(1-cosĀ²Ī±)Ā²
= 8cosāµĪ± - 10cosĀ³Ī± + 3cosĪ± ā
6cosĪ±
+ 6cosĀ³Ī± + 8cos -16cosĀ³Ī± +8cosāµĪ±,
Cos 5Ī± =
16cosāµ Ī± - 20cosĀ³ Ī± + 5cosĪ±
65)
Cos 5Ī±
= cos(2Ī± +3Ī±) = cos(2 Ī±)cos(3 Ī±) - sin(2 Ī±)sin(3 Ī±)
= (cosĀ²Ī± āsinĀ²Ī±) (cosĀ³Ī± ā3sinĀ²Ī± cos Ī±) ā 2sin Ī± cos Ī± (3sin Ī± cosĀ² Ī± āsinĀ³ Ī±)
= cosāµĪ± ā 3sinĀ² Ī± cosĀ³ Ī± ā sinĀ² Ī± cosĀ³ Ī± + 3sinā“Ī± cos Ī± ā 6sinĀ² Ī± cosĀ³ Ī± + 2sinā“ Ī± cos Ī±
= 5sinā“ Ī± cos Ī± ā 10sinĀ² Ī± cosĀ³ Ī± + cosāµĪ±
= 5(1ācosĀ² Ī±)Ā²cos Ī± ā 10(1ācosĀ² Ī±)cosĀ³ Ī± + cosāµ Ī±
= 5cos Ī± (cosā“ Ī± ā2cosĀ² Ī± +1) ā 10cosĀ³ Ī± (1ācosĀ² Ī±) + cosāµ Ī±
= 5cosāµ Ī± ā 10cosĀ³ Ī± + 5cos Ī± ā 10cosĀ³ Ī± + 10cosāµ Ī± + cosāµ Ī±
= (cosĀ²Ī± āsinĀ²Ī±) (cosĀ³Ī± ā3sinĀ²Ī± cos Ī±) ā 2sin Ī± cos Ī± (3sin Ī± cosĀ² Ī± āsinĀ³ Ī±)
= cosāµĪ± ā 3sinĀ² Ī± cosĀ³ Ī± ā sinĀ² Ī± cosĀ³ Ī± + 3sinā“Ī± cos Ī± ā 6sinĀ² Ī± cosĀ³ Ī± + 2sinā“ Ī± cos Ī±
= 5sinā“ Ī± cos Ī± ā 10sinĀ² Ī± cosĀ³ Ī± + cosāµĪ±
= 5(1ācosĀ² Ī±)Ā²cos Ī± ā 10(1ācosĀ² Ī±)cosĀ³ Ī± + cosāµ Ī±
= 5cos Ī± (cosā“ Ī± ā2cosĀ² Ī± +1) ā 10cosĀ³ Ī± (1ācosĀ² Ī±) + cosāµ Ī±
= 5cosāµ Ī± ā 10cosĀ³ Ī± + 5cos Ī± ā 10cosĀ³ Ī± + 10cosāµ Ī± + cosāµ Ī±
Cos 5Ī± = 16cosāµĪ± ā 20cosĀ³ Ī±
+ 5cos Ī± (terbukti)
66)
Tan 5Ī± = sin 5Ī±/cos 5Ī±
= (5 sin Ī± - 20 sinĀ³ Ī± + 16 sināµ Ī±)/(16cosāµ Ī± - 20cosĀ³ Ī± + 5cosĪ±)
Tan 5Ī± = (5tan Ī± - 10 tanĀ³Ī± + tanāµĪ±)/ (1-10tanĀ²Ī± + 5tanā“Ī±) (Terbukti)
= (5 sin Ī± - 20 sinĀ³ Ī± + 16 sināµ Ī±)/(16cosāµ Ī± - 20cosĀ³ Ī± + 5cosĪ±)
Tan 5Ī± = (5tan Ī± - 10 tanĀ³Ī± + tanāµĪ±)/ (1-10tanĀ²Ī± + 5tanā“Ī±) (Terbukti)
67)
Sin (6Ī±) = sin (2 Ā· 3Ī±) = 2sin (3Ī±) cos (3Ī±)
sin (3Ī±) = sin (2Ī± + Ī±) = sin (2Ī±) cos (Ī±) + cos (2Ī±) sin (Ī±)
cos (3Ī±) = cos (2Ī± + Ī±) = cos (2Ī±) cos (Ī±) - sin (2Ī±) cos (Ī±)
sin (2Ī±) = 2sin (Ī±) cos (Ī±)
cos (2Ī±) = cosĀ² (Ī±) - sinĀ² (Ī±)
jadi, kita mendapatkan
2sin (3Ī±) cos (3Ī±) = 2 (sin (2Ī±) cos (Ī±) + cos (2Ī±) sin (Ī±)) (cos (2Ī±) cos (Ī±) - sin (2Ī±) cos (Ī±))
= 2 (2sin (Ī±) cosĀ² (Ī±) + cosĀ² (Ī±) sin (Ī±) - sinĀ³ (Ī±)) (cosĀ³ (Ī±) -sin(Ī±) cosĀ²(Ī±) - 2sin (Ī±) cosĀ²(Ī±))
= 2 (3sin (Ī±) cosĀ²(Ī±) - sinĀ³ (Ī±)) (cosĀ³ (Ī±) - 3sin (Ī±) cosĀ²(Ī±))
= 2 [3sin (Ī±) cosāµ (Ī±) -9sinĀ² (Ī±) cosāµ (Ī±) -3sinā“ (Ī±) cosĀ² (Ī±) + 3sinā“ (Ī±) cos (Ī±)]
Sin 6Ī± = 6sin (Ī±) cos (Ī±) [cosā“ (Ī±) - 3sin (Ī±) cosā“ (Ī±) - sinĀ³ (Ī±) cos (Ī±) + sinĀ³ (Ī±)] (terbukti)
sin (3Ī±) = sin (2Ī± + Ī±) = sin (2Ī±) cos (Ī±) + cos (2Ī±) sin (Ī±)
cos (3Ī±) = cos (2Ī± + Ī±) = cos (2Ī±) cos (Ī±) - sin (2Ī±) cos (Ī±)
sin (2Ī±) = 2sin (Ī±) cos (Ī±)
cos (2Ī±) = cosĀ² (Ī±) - sinĀ² (Ī±)
jadi, kita mendapatkan
2sin (3Ī±) cos (3Ī±) = 2 (sin (2Ī±) cos (Ī±) + cos (2Ī±) sin (Ī±)) (cos (2Ī±) cos (Ī±) - sin (2Ī±) cos (Ī±))
= 2 (2sin (Ī±) cosĀ² (Ī±) + cosĀ² (Ī±) sin (Ī±) - sinĀ³ (Ī±)) (cosĀ³ (Ī±) -sin(Ī±) cosĀ²(Ī±) - 2sin (Ī±) cosĀ²(Ī±))
= 2 (3sin (Ī±) cosĀ²(Ī±) - sinĀ³ (Ī±)) (cosĀ³ (Ī±) - 3sin (Ī±) cosĀ²(Ī±))
= 2 [3sin (Ī±) cosāµ (Ī±) -9sinĀ² (Ī±) cosāµ (Ī±) -3sinā“ (Ī±) cosĀ² (Ī±) + 3sinā“ (Ī±) cos (Ī±)]
Sin 6Ī± = 6sin (Ī±) cos (Ī±) [cosā“ (Ī±) - 3sin (Ī±) cosā“ (Ī±) - sinĀ³ (Ī±) cos (Ī±) + sinĀ³ (Ī±)] (terbukti)
68)
Sin(6Ī±) = sin(3Ī± +
3Ī±) = 2sin(3Ī±)cos(3Ī±)
sin(3Ī±) = sin(2Ī± + Ī±) = sin(2Ī±)cos(Ī±) + cos(2Ī±)sin(Ī±)
cos(3Ī±) = cos(2Ī± + Ī±) = cos(2Ī±)cos(Ī±) - sin(2Ī±)sin(Ī±)
sin(2Ī±) = 2sin(Ī±)cos(Ī±)
cos(2Ī±) = cosĀ²(Ī±) - sinĀ²(Ī±)
sehingga sin(3Ī±) = 2sin(Ī±)cosĀ²(Ī±) + sin(Ī±)cosĀ²(Ī±) - sinĀ³(Ī±) = 3sin(Ī±)cosĀ²(Ī±) - sinĀ³(Ī±).
dan cos(3Ī±) = cosĀ³(Ī±) - cos(Ī±)sinĀ²(Ī±) - 2cos(Ī±)sinĀ²(Ī±) = cosĀ³(Ī±) - 3cos(Ī±)sinĀ²(Ī±).
sin(6Ī±) = 2(3sin(Ī±)cosĀ²(Ī±) - sinĀ³(Ī±))(cosĀ³(Ī±) - 3cos(Ī±)sinĀ²(Ī±))
= 2(3sin(Ī±)cosāµ(Ī±) - 9(cos(Ī±)sin(Ī±))Ā³ - (sin(Ī±)cos(Ī±))Ā³ + 3cos(Ī±)sināµ(Ī±))
Sin 6Ī±= 6sin(Ī±)cosāµ(Ī±) - 20(cos(Ī±)sin(Ī±))Ā³ + 6cos(Ī±)sināµ(Ī±)
sin(3Ī±) = sin(2Ī± + Ī±) = sin(2Ī±)cos(Ī±) + cos(2Ī±)sin(Ī±)
cos(3Ī±) = cos(2Ī± + Ī±) = cos(2Ī±)cos(Ī±) - sin(2Ī±)sin(Ī±)
sin(2Ī±) = 2sin(Ī±)cos(Ī±)
cos(2Ī±) = cosĀ²(Ī±) - sinĀ²(Ī±)
sehingga sin(3Ī±) = 2sin(Ī±)cosĀ²(Ī±) + sin(Ī±)cosĀ²(Ī±) - sinĀ³(Ī±) = 3sin(Ī±)cosĀ²(Ī±) - sinĀ³(Ī±).
dan cos(3Ī±) = cosĀ³(Ī±) - cos(Ī±)sinĀ²(Ī±) - 2cos(Ī±)sinĀ²(Ī±) = cosĀ³(Ī±) - 3cos(Ī±)sinĀ²(Ī±).
sin(6Ī±) = 2(3sin(Ī±)cosĀ²(Ī±) - sinĀ³(Ī±))(cosĀ³(Ī±) - 3cos(Ī±)sinĀ²(Ī±))
= 2(3sin(Ī±)cosāµ(Ī±) - 9(cos(Ī±)sin(Ī±))Ā³ - (sin(Ī±)cos(Ī±))Ā³ + 3cos(Ī±)sināµ(Ī±))
Sin 6Ī±= 6sin(Ī±)cosāµ(Ī±) - 20(cos(Ī±)sin(Ī±))Ā³ + 6cos(Ī±)sināµ(Ī±)
69)
Sin 6Ī± = sin
(4Ī± +2Ī±) = sin4Ī±cos2Ī± + cos4Ī±sin2Ī± = 2sin2Ī±.cos2Ī±.cos2Ī± + (1 ā 2 sinĀ²2Ī±)sin2Ī± =
2sin2Ī± (1 ā sinĀ²2Ī±) + (1 ā 2sinĀ²2Ī±)sin2Ī± = 2 sin2Ī± ā 2 sinĀ³2Ī± + sin2Ī± ā 2sinĀ³2Ī±
= 3sin2Ī± ā 4sinĀ³2Ī±
70)
Cos 6Ī± = cos 3(2Ī±)
cos 3Ī±
= 4 cosĀ³Ī± - 3cos Ī±
cos 6Ī± = 4 [ (2 cosĀ² (Ī±) - 1)Ā³ - 3 ( 2 cosĀ²(Ī±) -1 ) ]
= 4 [ 2 cosĀ² (Ī±) - (1)Ā³ - 3( 2 cosĀ²(Ī±) )Ā² + 3 ( 2 cosĀ² (Ī±) ] - 6 cosĀ² (Ī±) + 3
= 4 [ 8 cosā¶ (Ī±) - 1 - 12 cosā“ (Ī±) + 6 cosĀ² (Ī±) ] - 6 cosĀ² (Ī±) + 3
= 32 cosā¶ (Ī±) - 4 - 48 cosā“ (Ī±) + 24 cosĀ² (Ī±) - 6 cosĀ² (Ī±) + 3
= 32 cosā¶ (Ī±) - 48 cosā“ (Ī±) + 18 cosĀ² (Ī±) - 1
Cos 6Ī± = 32 cosā¶ (Ī±) - 48 cosā“ (Ī±) + 18 cosĀ² (Ī±) - 1 (terbukti)
cos 6Ī± = 4 [ (2 cosĀ² (Ī±) - 1)Ā³ - 3 ( 2 cosĀ²(Ī±) -1 ) ]
= 4 [ 2 cosĀ² (Ī±) - (1)Ā³ - 3( 2 cosĀ²(Ī±) )Ā² + 3 ( 2 cosĀ² (Ī±) ] - 6 cosĀ² (Ī±) + 3
= 4 [ 8 cosā¶ (Ī±) - 1 - 12 cosā“ (Ī±) + 6 cosĀ² (Ī±) ] - 6 cosĀ² (Ī±) + 3
= 32 cosā¶ (Ī±) - 4 - 48 cosā“ (Ī±) + 24 cosĀ² (Ī±) - 6 cosĀ² (Ī±) + 3
= 32 cosā¶ (Ī±) - 48 cosā“ (Ī±) + 18 cosĀ² (Ī±) - 1
Cos 6Ī± = 32 cosā¶ (Ī±) - 48 cosā“ (Ī±) + 18 cosĀ² (Ī±) - 1 (terbukti)
71)
Cos 6Ī± = cos(4Ī±+2Ī±)
= cos4Ī± cos2Ī±-sin4Ī± sin2Ī±
= [2cosĀ²2Ī±-1][2cosĀ²Ī±-1]-[2sin2Ī±cos2Ī±][2sinĪ±cosĪ±]
= [2(2cosĀ²Ī±-1)Ā²-1][2cosĀ²Ī±-1]-[2*2sinĪ±cosĪ±(2cosĀ²Ī±-1)][2sinĪ±cosĪ±]
= [2(4cosā“Ī±-4cosĀ²Ī±+1)-1][2cosĀ²Ī±-1]-[8sinĀ²Ī±cosĀ²Ī±(2cosĀ²Ī±-1)]
= [8cosā“Ī±-8cosĀ²Ī±+1][2cosĀ²Ī±-1]-[8(1-cosĀ²Ī±)cosĀ²Ī±(2cosĀ²Ī±-1)]
= [16cosā¶Ī±-8cosā“Ī±-16cosā“Ī±+8cosĀ²Ī±+2cosĀ²Ī±-1]-[(8cosĀ²Ī±-8cosā“Ī±)(2cosĀ²Ī±-1)]
= [16cosā¶Ī±-24cosā“Ī±+10cosĀ²Ī±-1]-[16cosā“Ī±-8cosĀ²Ī±-16cosā¶Ī±+8cosā“Ī±]
= [16cosā¶Ī±-24cosā“Ī±+10cosĀ²Ī±-1]-[24cosā“Ī±-8cosĀ²Ī±-16cosā¶Ī±]
Cos 6Ī±= 32cosā¶Ī± - 48cosā“Ī± + 18cosĀ²Ī±-1 (terbukti)
= cos4Ī± cos2Ī±-sin4Ī± sin2Ī±
= [2cosĀ²2Ī±-1][2cosĀ²Ī±-1]-[2sin2Ī±cos2Ī±][2sinĪ±cosĪ±]
= [2(2cosĀ²Ī±-1)Ā²-1][2cosĀ²Ī±-1]-[2*2sinĪ±cosĪ±(2cosĀ²Ī±-1)][2sinĪ±cosĪ±]
= [2(4cosā“Ī±-4cosĀ²Ī±+1)-1][2cosĀ²Ī±-1]-[8sinĀ²Ī±cosĀ²Ī±(2cosĀ²Ī±-1)]
= [8cosā“Ī±-8cosĀ²Ī±+1][2cosĀ²Ī±-1]-[8(1-cosĀ²Ī±)cosĀ²Ī±(2cosĀ²Ī±-1)]
= [16cosā¶Ī±-8cosā“Ī±-16cosā“Ī±+8cosĀ²Ī±+2cosĀ²Ī±-1]-[(8cosĀ²Ī±-8cosā“Ī±)(2cosĀ²Ī±-1)]
= [16cosā¶Ī±-24cosā“Ī±+10cosĀ²Ī±-1]-[16cosā“Ī±-8cosĀ²Ī±-16cosā¶Ī±+8cosā“Ī±]
= [16cosā¶Ī±-24cosā“Ī±+10cosĀ²Ī±-1]-[24cosā“Ī±-8cosĀ²Ī±-16cosā¶Ī±]
Cos 6Ī±= 32cosā¶Ī± - 48cosā“Ī± + 18cosĀ²Ī±-1 (terbukti)
72)
cos 6Ī± = 2cosĀ²Ī±
3Ī± - 1
= 2(4cosĀ³Ī± - 3cosĪ±)Ā² - 1
= 2(16 cosā¶Ī± + 9cosĀ²Ī± - 24cosĀ²Ī±) - 1
Cos 6Ī±= 32cosā¶Ī± - 48cosā“Ī± + 18cosĀ²Ī±-1 (terbukti)
= 2(4cosĀ³Ī± - 3cosĪ±)Ā² - 1
= 2(16 cosā¶Ī± + 9cosĀ²Ī± - 24cosĀ²Ī±) - 1
Cos 6Ī±= 32cosā¶Ī± - 48cosā“Ī± + 18cosĀ²Ī±-1 (terbukti)
73)
Sin
8Ī± = sin 2(4Ī±) kita tahu bahwa sin 2Ī± = 2sinĪ±.cosĪ±
Sin 2(4Ī±) = 2 sin 4Ī±.cos 4Ī± sehingga
Sin 4Ī± = sin 2(2Ī±) = 2sin2Ī±.cos2Ī± maka
Sin 8Ī± = 2{2sin 2Ī±cos 2Ī±}cos 4Ī± = 4
sin 2Ī±cos 2Ī±cos 4Ī±
Memasukkan sin 2Ī± = 2sinĪ±cosĪ± ke dalam
persamaan
= 2 sin 4Ī±cos4Ī± = 2(2sin2Ī±.cos2Ī±)cos
4Ī± = 4 sin 2Ī±cos 2Ī±cos 4Ī± = 4(2sinĪ±.cosĪ±)cos 2Ī±cos 4Ī±
Sin 8Ī± = 8sinĪ±.cosĪ±.cos2Ī±.cos4Ī±
74)
Cos 8Ī± = cos(4Ī±+4Ī±) = cos 4Ī±.cos4Ī± ā sin 4Ī±.sin4Ī± = cosĀ²4Ī± ā sinĀ²4Ī±
75)
Cos 8Ī± = cosĀ²4Ī± - sinĀ²4Ī± = (1-sinĀ²4Ī±) ā sinĀ²4Ī± = 1 ā 2 sinĀ²4Ī±
76)
Cos 8Ī± = cosĀ²4Ī± ā sinĀ²4Ī± = cosĀ²4Ī± ā (1 ā cosĀ²4Ī±) = 2cosĀ²4Ī± ā 1.
77)
buktikan
identitas berikut :sec 2Ī± = ( cot Ī± + tan Ī± ) /
(cot Ī± - tan Ī±)!
= [ (cos Ī± /
sin Ī± )+( sin Ī± / cos Ī± ) ] / [(cos Ī± /
sin Ī±)-(sin Ī± /cos Ī±) ]
= [cosĀ²( Ī± ) + sinĀ²( Ī±) ] / [ cosĀ²( Ī± ) - sinĀ²( Ī±) ]
= 1 / ( cos (2Ī±))
= sec (2Ī±)
= [cosĀ²( Ī± ) + sinĀ²( Ī±) ] / [ cosĀ²( Ī± ) - sinĀ²( Ī±) ]
= 1 / ( cos (2Ī±))
= sec (2Ī±)
78)
Tan 2Ī±= (2 tan Ī±)/(1-tanĀ²Ī±)
79)
Tan 3Ī± =
(3tan Ī± - tanĀ³Ī±)/(1 - 3tanĀ²Ī±)
80)
Tan 4É = {4tanĪ± ā 4 tanĀ³Ī±}/{1-6tanĀ²Ī± + tanā“Ī±}
81)
Tan 5Ī± = (5tan Ī± -
10 tanĀ³Ī± + tanāµĪ±)/ (1-10tanĀ²Ī± + 5tanā“Ī±)
82)
Tan 6Ī± = (6tanĪ± - 20tanĀ³Ī± + 6tanāµĪ±)/ (1-15tanĀ²Ī± + 15tanā“Ī± - tanā¶Ī±)
83)
Tan 7Ī± = (7tanĪ± - 35tanĀ³Ī± + 21tanāµĪ±- tanā·Ī±)/(1-21tanĀ²Ī± + 35tanā“Ī± - 7tanā¶Ī±)
84)
Tan 8Ī± = (8tanĪ± - 56tanĀ³Ī± + 56tanāµĪ± -
8tanā·Ī±)/(1-28tanĀ²Ī± + 70tanā“Ī± -
28tanā¶Ī± + tanāøĪ±)
85)
Tan 9Ī± = (9tanĪ± - 84tanĀ³Ī± + 126tanāµĪ± -36tanā·Ī± + tanā¹Ī±)/(1-
36tanĀ²Ī± + 126tanā“Ī± - 84tanā¶Ī± + 9tanāøĪ±)
86)
Tan 10Ī± = {10tanĪ± ā 120tanĀ³Ī± +252tanāµĪ± ā 120tanā·Ī± + 10 tanā¹Ī±}/{1 -45tanĀ²Ī± + 210tanā“Ī± ā 120tanā¶Ī± +45tanāøĪ± ā
tanĀ¹Ā°Ī±}
87)
Buktikan bahwa {1 ā
tanĀ²Ī±}/{1+tanĀ²Ī±} = cos 2Ī±
Ruas kiri = {1 ā tanĀ²Ī±}/{1+tanĀ²Ī±} =
{1-(sinĀ²Ī±/cosĀ²Ī±)}/{1+(sinĀ²Ī±/cosĀ²Ī±)}
= {(cosĀ²Ī±/cosĀ²Ī±)-(sinĀ²Ī±/cosĀ²Ī±)}/{(cosĀ²Ī±/cosĀ²Ī±) +
(sinĀ²Ī±/cosĀ²Ī±)}
= (cosĀ²Ī± ā sinĀ²Ī±) / (cosĀ²Ī± + sinĀ²Ī±) = cos 2Ī±/1 =
cos 2Ī± ruas kanan
88)
Buktikan bahwa sin 2Ī± =
(2cot Ī±)/(1+cotĀ²Ī±)
Ruas kanan = (2cot Ī±)/(1+cotĀ²Ī±) = {2(cos
Ī±/sinĪ±)}/{1+(cosĀ²Ī±/sinĀ²Ī±)}
= (2sin Ī±.cos Ī±)/(sinĀ²Ī±+cosĀ²Ī±) = 2 sinĪ±.cosĪ± = sin 2Ī±
89)
Buktikan
bahwa {1 ā tanĀ²Ī±}/{1+tanĀ²Ī±} = cos 2Ī±
Ruas kiri = {1 ā
tanĀ²Ī±}/{1+tanĀ²Ī±} = {1-(sinĀ²Ī±/cosĀ²Ī±)}/{1+(sinĀ²Ī±/cosĀ²Ī±)}
=
{(cosĀ²Ī±/cosĀ²Ī±)-(sinĀ²Ī±/cosĀ²Ī±)}/{(cosĀ²Ī±/cosĀ²Ī±) + (sinĀ²Ī±/cosĀ²Ī±)}
= (cosĀ²Ī± ā sinĀ²Ī±) / (cosĀ²Ī±
+ sinĀ²Ī±) = cos 2Ī±/1 = cos 2Ī± ruas kanan
90)
Buktikan
bahwa sin 2Ī± = (2cot Ī±)/(1+cotĀ²Ī±)
Ruas kanan = (2cot
Ī±)/(1+cotĀ²Ī±) = {2(cos Ī±/sinĪ±)}/{1+(cosĀ²Ī±/sinĀ²Ī±)}
= (2sin Ī±.cos
Ī±)/(sinĀ²Ī±+cosĀ²Ī±) = 2 sinĪ±.cosĪ± = 2 sin Ī±
91)
Cos
2Ī± = 2cosĀ²Ī± ā 1
92)
Cos
3Ī± = 4cosĀ³Ī± ā 3 cosĪ±
93)
Cos
4Ī± = 8 cosā“Ī± ā 8 cosĀ²Ī± + 1
94)
Cos 5Ī±
= 16cosāµĪ± ā 20 cosĀ³Ī± + 5 cos Ī±.
95)
Cos
6Ī± = 32cosā¶Ī± ā 48 cosā“Ī± + 18 cosĀ²Ī± ā 1.
96)
Cos
7Ī± = 64cosā·Ī± ā 112 cosāµĪ± + 56 cosĀ³Ī± ā 7cosĪ±
97)
Cos
8Ī± = 128cosāøĪ± ā 256 cosā¶Ī± + 160 cosā“Ī± ā 32cosĀ²Ī±
+ 1
98)
Cos 9Ī± = 256cosā¹Ī± ā 576 cosā·Ī±
+ 432 cosāµĪ± ā 120cosĀ³Ī± + 9cosĪ±
99)
Cos 10Ī± = 512cosĀ¹įµĪ± ā 1280 cosāøĪ± + 1120cosā¶Ī± ā 400cosā“Ī± + 50
cosĀ²Ī± ā 1
100) Sin 2Ī± = 2sin Ī± cosĪ±
101) Sin 3Ī± = 3sinĪ± ā 4sinĀ³Ī±
102) Sin 4Ī± = 2sin 2Ī± (1-2sinĀ²Ī±) = 2 sin 2Ī± ā 4sin2Ī±
sinĀ²Ī± = 4sinĪ±.cosĪ± ā 8sinĀ³Ī±cosĪ±
103) Sin 5Ī± = 5sin Ī± ā 20 sinĀ³Ī± + 16sināµĪ±
104) Sin 6Ī± = 3sin 2Ī± ā 4sinĀ³2Ī± =
6sinĪ±.cosĪ±-32sinĀ³Ī±cosĪ±+32sināµĪ±cosĪ±
105) Sin 7Ī± = 7sin Ī± ā 56sinĀ³Ī± + 112sināµĪ± ā 64 sinā·Ī±.
106) Sin 8Ī±=
8sinĪ±cosĪ±-80sinĀ³Ī±cosĪ±+192sināµĪ±cosĪ±-128sinā·Ī±cosĪ±
107) Sin 9Ī± = 9sinĪ±-120sinĀ³Ī±+432sināµĪ±-576sinā·Ī±+256sinā¹Ī±
108) Sin 10Ī± = 10sinĪ±cosĪ±-160sinĀ³Ī±cosĪ±+672sināµĪ±cosĪ±-1024sinā·Ī±cosĪ±+5125sinā¹Ī±cosĪ±
109) sin(3Ī±) = 3
sin(Ī±) - 4 sinĀ³(Ī±),
110) cos(3Ī±) = 4 cosĀ³(Ī±) - 3 cos(Ī±),
111) tan(3Ī±) = [3 tan(Ī±)-tanĀ³(Ī±)]/[1-3 tanĀ²(Ī±)]
112) sin(4Ī±) = 4 sin(Ī±)cos(Ī±)[2 cosĀ²(Ī±)-1],
113) cos(4Ī±) = 8 cosā“(Ī±) - 8 cosĀ²(Ī±) + 1.
114) sin(5Ī±) = 5 sin(Ī±) - 20 sinĀ³(Ī±) + 16 sināµ(Ī±),
115) cos(5Ī±) = 16 cosāµ(Ī±)
- 20 cosĀ³(Ī±) + 5 cos(Ī±).
116) sin(6Ī±) = 2 sin(Ī±)cos(Ī±)[16
cosā“(Ī±) - 16 cosĀ²(Ī±) + 3],
117) cos(6Ī±) = 32 cosā¶(Ī±) - 48 cosā“(Ī±) + 18 cosĀ²(Ī±) - 1.
sin(nx) = 2 sin([n-1]x)cos(x) - sin([n-2]x),
cos(nx) = 2 cos([n-1]x)cos(x) -
cos([n-2]x),
tan(nx)
= (tan[(n-1)x]+tan[x])/(1-tan[(n-1)x]tan[x]) karena tan nx = tan{(n-1)x+x} jadi
tan nx = tan{(n-1)x+x}={tan(n-1)x+tanx}/{1-tan(n-1)x.tanx} jika tan (n-1)x =
H/k maka tan nx = {(h/k)+tan x}/{1-(H/k)tanx} maka
tan (nx) = {H+tanx.k}/{k-Htan
x}
lanjutan Sudut Ganda atau rangkap part 2 klik disini
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