Proof of the product and sum formulas
a) | sin | = | ½[sin ( |
b) | cos | = | ½[sin ( |
c) | cos | = | ½[cos ( |
d) | sin | = | −½[cos ( |
These formulas are also derived from the sum and difference formulas. To derive (a), write
sin ( | = | sin |
sin ( | = | sin |
and add vertically. The last terms in each line will cancel:
sin (
+ β) + sin (
− β) = 2 sin
cos β.
Therefore, on exchanging sides,
2 sin
cos β = sin (
+ β) + sin (
− β),
so that
sin
cos β = ½[sin (
+ β) + sin (
− β)].
This is the identity (a)).
Formula (b) is derived in exactly the same manner, only instead of adding, subtract sin (
− β) from sin (
+ β).
Formulas (c) and (d) are derived similarly. To derive (c), write
cos (
+ β) = cos
cos β − sin
sin β,
cos (
− β) = cos
cos β + sin
sin β,
and add. To derive (d), subtract.
Let us derive (d). On subtracting, the first terms on the right will cancel. We will have
cos (
+ β) − cos (
− β) = −2 sin
sin β.
Therefore, on solving for sin
sin β,
sin
sin β = −½[cos (
+ β) − cos (
− β)].
e) | sin A + sin B | = | 2 sin ½ (A + B) cos ½ (A − B) |
f) | sin A − sin B | = | 2 sin ½ (A − B) cos ½ (A + B) |
g) | cos A + cos B | = | 2 cos ½ (A + B) cos ½ (A − B) |
h) | cos A − cos B | = | −2 sin ½ (A + B) sin ½ (A − B) |
The formulas (e), (f), (g), (h) are derived from (a), (b), (c), (d)respectively; that is, (e) comes from (a), (f) comes from (b), and so on.
To derive (e), exchange sides in (a):
½[sin (
+ β) + sin (
− β)] = sin
cos β,
so that
= | A | |||||
and | ||||||
= | B. . . . . . . . . . . . . . . .(2) |
The left-hand side of line (1) then becomes
sin A + sin B.
This is now the left-hand side of (e), which is what we are trying to prove.
On adding them, 2
= A + B,
so that
On subtracting those two equations, 2β = A − B,
so that
β = ½(A − B).
On the right−hand side of line (1), substitute those expressions for
and β. Line (1) then becomes
sin A + sin B = 2 sin ½(A + B) cos ½(A − B).
This is the identity (e).
Read it as follows:
"sin A + sin B equals twice the sine of half their sum
times the cosine of half their difference."
times the cosine of half their difference."
Identities (f), (g), and (h) are derived in exactly the same manner from(b), (c), and (d) respectively.
Rumus-rumus seperti itu sangat kita perlukan untuk menghadapi soal-soal. Terutama soal-soal ujian, ujian nasional, snmptn atau yang lainnya(terutama skripshit saia). Sebenarnya dari mana rumus tersebut berasal. Perhatikan hal di bawah ini
Rumus tersebut sebenarnya hanya berasal dari rumus cosinus jumlah dan cosinus selisih sudut.
Kedua rumus atau kedua persamaan tersebut kita jumlahkan. didapatkan :
Menjadi
rumus-rumus pada trigonometri sebenarnya yang perlu dihafal hanya rumus jumlah dan selisih sudutnya saja. Karena rumus yang lain dengan mudah kita bisa mendapatkannya. Misalkan saja untuk mencari rumus perkalian : 
kita tinggal mengurangkan kedua persamaan tersebut. dan nantinya akan dengan mudah didapatkan rumusnya. Tidak perlu menghafalnya. Hanya menghafal rumus jumlah dan selisih sudut untuk sin dan cos saja.
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