### Proof 1 - Using the Unit Circle

We construct angles `BOA = alpha` and `AOP = beta` as shown.

Next, we drop a perpendicular from P to the x-axis at T. Point C is the intersection of OA and PT.

We then construct line PR perpendicular to OA.

Finally, we drop a perpendicular from R to the x-axis at S, and another from R to PT at Q, as shown.

We note the following:

(1) `/_TPR = alpha` since triangles OTC and PRC are similar. (`/_OTC = /_PRC = 90°`, and `/_OCT = /_PCR = 90°- alpha`.)

(2) Length |QT| = |RS|

(3) sin (α + β) = |PT| = |PQ| + |QT| = |PQ| + |RS|

(4) |PR| = sin (β)

(5) In triangle PQR, |PQ| = |PR| cos (α)

(6) So from (4) and (5), |PQ| = sin (β) cos (α).

(7) |OR| = cos (β)

(8) In triangle OSR, |RS| = |OR| sin (α)

(9) So from (7) and (8), |RS| = cos (β) sin (α)

(10) Thus from (3), (6) and (9), we have proved:

sin (α + β) = sin (β) cos (α) + cos (β) sin (α)

Rearranging gives:

sin (α + β) = sin (α) cos (β) + cos (α) sin (β)

(11) From Even and Odd Functions, we have: cos (−β) = cos( β) and sin (−β) = −sin(β)

(12) So replacing β with (−β), the identity in (10) becomes

sin (αβ) = sin α cos β − cos α sin β

[Thank you to David McIntosh for providing the outline of the above proof.]

#### The Cosine Proofs

We now need to prove

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

(13) |OT| = cos (α + β)

(14) In triangle ORS, we have: `cos alpha = |OS|/|OR|`.

(15) cos (β) = |OR|, from (7) above.

(16) From (14) and (15), we obtain `cos alpha cos beta= |OS|/|OR|xx|OR| = |OS|`.

(17) In triangle QPR, we have `sin alpha = |QR|/|PR|`.

(18) sin (β) = |PR|, from (4) above.

(19) From (17) and (18), we obtain `sin alpha sin beta= |QR|/|PR|xx|PR| = |QR|`.

(20) Now |OS| − |QR| = |OT|.

(21) So `cos alpha cos beta - sin alpha sin beta` ` = cos(alpha+beta)`.

(22) Rearranging, we have:

cos (α + β) = cos α cos β sin α sin β

(23) Once again, we replace β with (−β), and the identity in (22) becomes:

cos (αβ) = cos α cos β + sin α sin β