### Proof 2 - Using the Unit Circle

We will prove the cosine of the sum of two angles identity first, and then show that this result can be extended to all the other identities given.

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

We draw a circle with radius 1 unit, with point P on the circumference at (1, 0).

We draw an angle α from the centre with terminal point Q at (cos α, sin α), as shown. [Q is (cos α, sin α) because the hypotenuse is 1 unit.]

We extend this idea by drawing:

• The angle β with terminal points at Q (cos α, sin α) and R (cos (α + β), sin (α + β))
• The angle -β with terminal point at S (cos (-β), sin (-β))
• The lines PR and QS, which are equivalent in length.

Now, using the distance formula from Analytical Geometry, we have:

PR2 = (cos (α + β) − 1)2 + sin2(α + β)

= cos2(α + β) − 2cos (α + β) + 1 + sin2(α + β)

= 2 − 2cos (α + β)

[since sin2(α + β) + cos2(α + β) = 1]

Now using the distance formula on distance QS:

QS2 = (cos α − cos (-β))2 + (sin α − sin (-β))2

= cos2α − 2cos α cos(-β) + cos2(-β) + sin 2α − 2sin α sin(-β) + sin2(-β)

= 2 − 2cos α cos(-β) − 2sin α sin(-β)

[since sin2α + cos2α = 1 and sin2(-β) + cos2(-β) = 1]

= 2 − 2cos α cos β + 2sin α sin β

[since cos(-β) = cos β (cosine is an even function) and sin(-β) = -sin β (sine is an odd function - see Even and Odd Functions)]

Since PR = QS, we can equate the 2 distances we just found:

2 − 2cos (α + β) = 2 − 2cos α cos β + 2sin α sin β

Subtracting 2 from both sides and dividing throughout by -2, we obtain:

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

If we replace β with (-β), this identity becomes:

cos (αβ) = cos α cos β + sin α sin β

[since cos(-β) = cos β and sin(-β) = -sinβ]

### Sine of the Sum of Two Angles

We aim to prove that

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

Recall that (see phase shift)

`sin (θ) = cos (π/2− θ)`

If ` θ = α + β`, then we have:

`sin (α + β) = cos [π/2 − (α + β)]`

`= cos [π/2 − α − β]`

Next, we re-group the angles inside the cosine term, since we need this for the rest of the proof:

`cos [π/2 − α − β] = cos [(π/2 − α) − β]`

Using the cosine of the difference of 2 angles identity that we just found above [which said cos (αβ) = cos α cos β + sin α sin β], we have:

`cos [(π/2 − α) − β]= cos (π/2 − α) cos\ β + sin (π/2 − α)\ sin\ β`

`= sin\ α\ cos\ β + cos\ α\ sin\ β`

[Since `cos (π/2 − α) = sin\ α`; and `sin (π/2 − α) = cos\ α`.]

Therefore:

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

Replacing β with (−β), this identity becomes (because of Even and Odd Functions):

sin (αβ) = sin α cos β − cos α sin β

We have proved the 4 identities involving sine and cosine of the sum and difference of two angles.

#### Summary:

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

sin (αβ) = sin α cos β − cos α sin β

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

cos (αβ) = cos α cos β + sin α sin β