Proof of the product and sum formulas

Products as sums

 a) sin cos β = ½[sin ( + β) + sin ( − β)] b) cos sin β = ½[sin ( + β) − sin ( − β)] c) cos cos β = ½[cos ( + β) + cos ( − β)] d) sin sin β = −½[cos ( + β) − cos ( − β)]

Proof

These formulas are also derived from the sum and difference formulas.  To derive (a), write

 sin ( + β) = sin cos β + cos sin β sin ( − β) = sin cos β − cos sin β

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)].

Sums as products

 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)

Proof

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

sin ( + β) + sin (β) = 2 sin cos β.  .  .  .  .  . (1)

Now put

 + β = A and − β = 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.

To complete the right−hand side of line (1), solve those simultaneous equations (2) for and β

On adding them,  2 = A + B,

so that

= ½(A + B).

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."

Identities (f), (g), and (h) are derived in exactly the same manner from (b), (c), and (d) respectively.

Trigonometric identities

Table of Contents | Home

Copyright © 2018 Lawrence Spector

Questions or comments?

E-mail:  themathpage@yandex.com

Private tutoring available.