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.]

proof

We extend this idea by drawing:

a. The angle β with terminal points at Q (cos α, sin α) and R (cos (α + β), sin (α + β))

b. The angle −β with terminal point at S (cos (−β), sin (−β))

c. The lines PR and QS, which are equivalent in length.

proof

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

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

= cos2(α + β) − 2 cos (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

= cos2 α − 2 cos 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) = −sinβ (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 (α − β) = 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 (α − β) = 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 (α − β) = 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 (α − β) = sin α cos β − cos α sin β

cos (α+β) = cos α cos β − sin α sin β

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

Get the Daily Math Tweet!
IntMath on Twitter