Proof of the product and sum formulas
Products as sums
These formulas are also derived from the sum and difference formulas. To derive (a), write
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),
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)].
Sums as products
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,
sin ( + β) + sin ( − β) = 2 sin cos β. . . . . . (1)
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,
= ½(A + B).
On subtracting those two equations, 2β = A − B,
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
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