Sum and di erence formulae for sine and cosinekws006/Precalculus/5.1_Sum_and...Part 5, Trigonometry...

Elementary FunctionsPart 5, Trigonometry

Lecture 5.1a, Sum and Difference Formulas

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 28

Sum and difference formulae for sine and cosine

Consider angles α and β with α > β. These angles identify points on theunit circle, P (cosα, sinα) and Q(cosβ, sinβ).



The distance between P and Q is, by the Pythagorean theorem, thesquare root of

d(PQ)2 = (cosα− cosβ)2 + (sinα− sinβ)2.



We can expand this out, algebraically, and get

d(PQ)2 = (cos2 α−2 cosα cosβ+cos2 β)+(sin2 α−2 sinα sinβ+sin2 β).

which we can rewrite (using the Pythagorean identity) as

d(PQ)2 = 2− 2 cosα cosβ − 2 sinα sinβ



Let’s rotate this picture clockwise through the angle −β so that the pointQ becomes Q′(1, 0) lying on the x-axis. The point P rotates into thepoint P ′(cos(α− β), sin(α− β))The distance from P ′ to Q′ is the same as the distance from P to Q.





The distance from P ′ to Q′ is the square root of

d(P ′Q′)2 =

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

= 2− 2 cos(α− β).



Since the line segments PQ and P ′Q′ are congruent, then we know thatd(PQ)2 = d(P ′Q′)2.



In mathematics, if we arrive at the same value through two differentcomputations, we always have valuable information. Here we can equated(PQ)2 and d(P ′Q′)2 and simplify.



D2 = 2− 2 cosα cosβ − 2 sinα sinβ

D2 = 2− 2 cos(α− β)2Smith (SHSU) Elementary Functions 2013 10 / 28


So2− 2 cosα cosβ − 2 sinα sinβ = 2− 2 cos(α− β)2.

We may divide both sides by 2 and solve for cos(α− β).

cos(α− β) = cosα cosβ + sinα sinβ

This is an important result.



We create a more memorable form of this equation if we replace β by −βand use the fact that sine is an odd function while cosine is an evenfunction.

Write cos(α+ β) = cos(α− (−β)) and replace β by −β in our equationfor cos(α− β) to see that

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

This is an equation we want to record and use on a regular basis.

cos(α+ β) = cosα cosβ − sinα sinβ (1)



If we note that sin(θ) = cos(π

2− θ) (since sine and cosine are

complementary functions!) then we can write

sin(α+ β) = cos(π

2− (α+ β)) = cos((

π

2− α)− β).

Using the sum-of-angles formula above, with regards to the anglesπ

2− α

and β, we have

sin(α+ β) = cos((π

2− α)− β)

= cos(π

2− α) cosβ + sin(

π

2− α) sinβ = sinα cosβ + cosα sinβ.



Therefore we have a sum-of-angles equation for the sine function:

sin(α+ β) = sinα cosβ + cosα sinβ (2)

If we need, we may replace β by −β to create a difference equation:

sin(α− β) = sinα cosβ − cosα sinβ



I don’t memorize these identities.

(That’s one reason I’ve been attracted to mathematics – if oneunderstands the math, one doesn’t need to memorize!)

In the undergraduate classes that I teach at Sam Houston State University,I will provide students with the sum of angle formulas when they areneeded.



Here they are again:

cos(α+ β) = cosα cosβ − sinα sinβ

sin(α+ β) = sinα cosβ + cosα sinβ



Here they are as difference of angles

cos(α−β) = cosα cosβ+sinα sinβ

sin(α−β) = sinα cosβ− cosα sinβ


Sum and Difference Formulas

In the next presentation, we will look at some applications of these sumand difference formulas.

(End)


Elementary FunctionsPart 5, Trigonometry

Lecture 5.1b, Applications of the Sum and Difference Formulas

Dr. Ken W. Smith

Sam Houston State University

2013


A reduction formula

Let P (cos θ, sin θ) be a point on the terminal side of angle θ.

Any point (a, b) on the line←→OP satisfies the equations

cos θ = a√a2+b2

, sin θ = b√a2+b2

.


A reduction formula

By our sum-of-angles formulas,

sin(x+ θ) = sinx cos θ + cosx sin θ.

If we replace cos θ and sin θ by a√a2+b2

and sin θ = b√a2+b2

, we get

sin(x+ θ) = sinx a√a2+b2

+ cosx b√a2+b2

= a sinx+b cosx√a2+b2

.

We clear the denominators by multiplying all sides by√a2 + b2.

a sinx+ b cosx =√a2 + b2 sin(x+ θ). (3)



a sinx+ b cosx =√a2 + b2 sin(x+ θ).

This formula is useful for changing a linear combination of sine and cosinefunctions into just a sine function.For example, the function f(x) = sinx+

√3 cosx can be rewritten as

sinx+√3 cosx =

√12 +

√32sin(x+ θ)

where θ is the angle between the x-axis and the line from the origin to thepoint (1,

√3). Since θ = π/3 in this problem, and since√

12 +√32=√1 + 3 = 2, we have

sinx+√3 cosx = 2 sin(x+ π/3).


Sum and difference formulas for tangent

We know that tan(α+ β) = sin(α+β)cos(α+β) so we may use our sum-of-angle

formulas to create a formula for tan(α+ β).A first pass gives

tan(α+ β) = sin(α+β)cos(α+β) =

sinα cosβ+cosα sinβcosα cosβ−sinα sinβ

But we would really like a sum-of-angles formula for tangent that is interms of tanα and tanβ.So let’s divide both numerator and denominator by cosα cosβ.

tan(α+ β) =tanα+ tanβ

1− tanα tanβ. (4)


Sum and difference formulae for tangent

tan(α+ β) =tanα+ tanβ

1− tanα tanβ.

What if we wanted an equation for tan(α− β)?

Replace β by −β and use the fact that tangent is an odd function toobtain

tan(α− β) = tanα− tanβ

1 + tanα tanβ.


Some worked problems

Compute the exact value of cos 75◦.

Solution.Using the sum of angles formula for cosine and the fact that75◦ = 45◦ + 30◦, we have

cos 75◦ = cos(45◦ + 30◦) = cos 45◦ cos 30◦ − sin 45◦ sin 30◦

=

√2

2

√3

2−√2

2

1

2=

√6−√2

4



Compute the exact value of sin 15◦.

Solution.

sin 15◦ = sin(45◦ − 30◦) = sin 45◦ cos 30◦ − cos 45◦ sin 30◦

=

√2

2

√3

2−√2

2

1

2=

√6−√2

4

This answer looks familiar! (Notice that since 75◦ and 15◦ arecomplementary angles so the cosine of one angle is the sine of the other!)



Find the tangent of 75◦.

Solution. We could separately compute the sine and cosine of 75◦ usingour sum-of-angle formulas (as we did on the last slides) or we could usethe sum-of-angles formula for tangent and write

tan 75◦ = tan(45◦ + 30◦) =tan 45◦ + tan 30◦

1− tan 30◦ tan 45◦=

1 +

√3

3

1− (

√3

3)(1)

.

Multiplying numerator and denominator by 3 gives

tan 75◦ =3 +√3

3−√3.

If we don’t like the square roots in the denominator, we can multiplynumerator and denominator by 3 +

√3 (the conjugate of the denominator)

and find that

tan 75◦ = (3 +√3

3−√3)(3 +√3

3 +√3) = (

9 + 6√3 + 3

9− 3) = (

12 + 6√3

6) = 2 +

√3 .Smith (SHSU) Elementary Functions 2013 27 / 28

Sum and Difference Formulas

In the next presentation, we will look at double angle and power reductionformulas.

(End)


Sum and di erence formulae for sine and cosinekws006/Precalculus/5.1_Sum_and...Part 5, Trigonometry...

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