Deriving the Sum and Difference Formulas for Sine and Cosine

In this discussion, we are going to derive the sum and difference identities, also known as the sum and difference formulas, for sine and cosine.

Here is an outline of this article:

Review of the Fundamental Trig Identities
Deriving the Difference Formula for Cosine from the Unit Circle
Deriving the Sum Formula for Cosine
Deriving the Sum Formula for Sine
Deriving the Difference Formula for Sine

It is beneficial to have a good algebra background before reading this article. However, I plan to show every step; so hopefully, you will find this article easy to follow regardless of your algebra background.

We will be using the fundamental Trigonometric Identities. If you need to review the fundamental trig identities, I will display them below.

Review of the Fundamental Trig Identities

Okay, let us begin!

Deriving the Difference Formula for Cosine

We begin with a unit circle.

Let the center of the circle be point O.

Since we are on the unit circle, the positive x-axis has terminal points (1,0) and we will call this point P.

The center of a circle to a point on the circle

Now draw an angle β anywhere you like. For this discussion, I will draw it somewhere in quadrant I.

The angle β has terminal points (cosβ, sinβ) and let’s call this point Q.

Next, draw an angle α wherever you like. For this discussion, I will draw it in quadrant II.

The angle α has terminal points (cosα, sinα) and let’s call this point R.

Now some of you might ask, “Since angle α is in the second quadrant, why did you put

(\cos α, \sin α)

instead of

(- \cos α, sine α) ? ”

If this is your question, you are just thinking about the output of cosine.

For example, in quadrant II, we have an angle

\frac{3 π}{4}

Cosine of

\cos (\frac{3 π}{4}) = - \frac{\sqrt{2}}{2}

You see, the output is negative,

but we say

\cos (\frac{3 π}{4})

NOT

- \cos (\frac{3 π}{4}) .

Hopefully this clears up any questions! Let us continue.

Now, if we begin at α and take away β, we have the angle

So, I draw the angle (α−β) and it has terminal points

and let’s call this point S.

Thus, from the positive x-axis to α-β we have the angle α-β.

In other words,

points S-O-P make the angle α-β.

Note that all points, from the center to the circle, has radius one since we are using the unit circle.

Now put on your thinking caps here!

If I begin at the positive x-axis, and take away β,

then from β to α, we also have the angle α-β.

In other words,

the points R-O-Q make the angle α-β.

Let me restate this :

We have two angles with α-β.

The points S-O-P make the angle (α-β).
and
the points R-O-Q make the angle (α-β).

This means that the distance from P to S is equal to the distance from Q to R.

So, we draw a line from P to S, and draw another line from Q to R.

Now we can use the distance formula that we learned from Precalculus Algebra.

Review:

The distance formula says that, given a point U(x1, y1) and V(x2, y2,), the distance between U and V is

Let:

P (x_{1}, y_{1}) = (1, 0)

S (x_{1}, y, 1) = [\cos (α - β), \sin (α - β)]

Q (x_{1}, y_{1}) = (\cos β, \sin β)

R (x_{2}, y_{2}) = (\cos α, \sin α)

Then:

d (P, S) = \sqrt{{[\cos (α - β) - 1]}^{2} + {[\sin (α - β) - 0]}^{2}}

Now let us take the distance between the points

P(1,0) and S(cos(α-β), sin(α-β))

and set it equal to the distance between the points

Q(cosβ, sinβ) and R(cosα, sinα).

This will give us the difference formula for cosine.

d (P, S) = d (Q, R)

\sqrt{{[\cos (α - β) - 1]}^{2} + {[\sin (α - β) - 0]}^{2}} = \sqrt{{[\cos α - \cos β]}^{2} + {[\sin α - \sin β]}^{2}}

• Square both sides

{[\cos (α - β) - 1]}^{2} + {[\sin (α - β) - 0]}^{2} = {[\cos α - \cos β]}^{2} + {[\sin α - \sin β]}^{2}

• Expand using FOIL

\cos^{2} (α - β) - 2 \cos (α - β) + 1 + \sin^{2} (α - β) = \cos^{2} α - 2 \cos α \cos β + \cos^{2} β + \sin^{2} α - 2 \sin α \sin β + \sin^{2} β

• Use sin²θ + cos²θ = 1

1 - 2 \cos (α - β) + 1 = 1 - 2 \cos α \cos β + 1 - 2 \sin α \sin β

• Combine like terms

2 - 2 \cos (α - β) = 2 - 2 \cos α \cos β - 2 \sin α \sin β

• Subtract 2 from both sides

- 2 \cos (α - β) = - 2 \cos α \cos β - 2 \sin α \sin β

• Divide both sides by −2

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

Deriving the Sum Formula for Cosine

Now we use the difference formula for cosine to find the sum formula for cosine.

• We know that:

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

\cos (α + β) = \cos [α - (- β)] = \cos α \cos (- β) + \sin α \sin (- β)

• Now, by the Even/Odd Identities:

\cos (- β) = \cos β and \sin (- β) = - \sin β

• Thus,

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

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

Deriving the Sum Formula for Sine

Now we use the cofunction identities and the difference formula for cosine to derive the sum formula for sine

• We know that:

\sin (θ) = \cos [\frac{π}{2} - θ] and \cos (θ) = \sin [\frac{π}{2} - θ]

\sin (α + β) = \cos [\frac{π}{2} - (α + β)] = \cos [(\frac{π}{2} - α) - β]

= \cos (\frac{π}{2} - α) \cos (β) + \sin (\frac{π}{2} - α) \sin (β)

• Thus,

\sin (α + β) = \cos (\frac{π}{2} - α) \cos (β) + \sin (\frac{π}{2} - α) \sin β

• Using the Cofunction Identities

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

Deriving the Difference Formula for Sine

Now we use the sum formula for sine to find the difference formula for sine.

• We know that:

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

\sin (α - β) = \sin [α + (- β)] = \sin α \cos (- β) + \cos α \sin (- β)

• Now, by the Even/Odd Identities:

\cos (- β) = \cos β and \sin (- β) = - \sin β

• Thus,

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

\sin (α - β) = \sin α \cos β - \cos α \sin β

We have derived the sum and difference formulas for sine and cosine!

Thank you for reading!

I hope that I can assist you on your math journey!

Sincerely,

MathAngel369

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Deriving the Sum and Difference Formulas for Sine and Cosine

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