Trigonometry by Michael Corral - HTML preview

1 + cos θ

1 + cos θ

2

2

2

The half-angle formulas are often used (e.g. in calculus) to replace a squared trigonometric

function by a nonsquared function, especially when 2 θ is used instead of θ.

3See p.331 in L.A. OSTROVSKY AND A.I.POTAPOV, Modulated Waves: Theory and Applications, Baltimore:

The Johns Hopkins University Press, 1999.

80

Chapter 3 • Identities

§3.3

By taking square roots, we can write the above formulas in an alternate form:

1 − cos θ

sin 1 θ

(3.31)

2

= ±

2

1 + cos θ

cos 1 θ

(3.32)

2

= ±

2

1 − cos θ

tan 1 θ

(3.33)

2

= ±

1 + cos θ

In the above form, the sign in front of the square root is determined by the quadrant in

which the angle 1 θ is located. For example, if θ

θ

2

= 300◦ then 12 = 150◦ is in QII. So in this

case cos 1 θ

θ

.

2

< 0 and hence we would have cos 12 = −

1 + cos θ

2

In formula (3.33), multiplying the numerator and denominator inside the square root by

(1 − cos θ) gives

1 − cos θ 1 − cos θ

(1 − cos θ)2

(1 − cos θ)2

1 − cos θ

tan 1 θ

.

2

= ±

·

= ±

= ±

= ±

1 + cos θ 1 − cos θ

1 − cos2 θ

sin2 θ

sin θ

But 1 − cos θ ≥ 0, and it turns out (see Exercise 10) that tan 1 θ and sin θ always have the 2

same sign. Thus, the minus sign in front of the last expression is not possible (since that

would switch the signs of tan 1 θ and sin θ), so we have:

2

1 − cos θ

tan 1 θ

(3.34)

2

=

sin θ

Multiplying the numerator and denominator in formula (3.34) by 1 + cos θ gives

1 − cos θ

1 + cos θ

1 − cos2 θ

sin2 θ

tan 1 θ

,

2

=

·

=

=

sin θ

1 + cos θ

sin θ (1 + cos θ)

sin θ (1 + cos θ)

so we also get:

sin θ

tan 1 θ

(3.35)

2

= 1 + cos θ

Taking reciprocals in formulas (3.34) and (3.35) gives:

sin θ

1 + cos θ

cot 1 θ

(3.36)

2

=

=

1 − cos θ

sin θ

Double-Angle and Half-Angle Formulas • Section 3.3

81

Example 3.15

2 sec θ

Prove the identity sec2 1 θ

.

2

= sec θ + 1

Solution: Since secant is the reciprocal of cosine, taking the reciprocal of formula (3.29) for cos2 1 θ

2

gives us

2

2

sec θ

2 sec θ

sec2 1 θ

.

2

=

=

·

=

1 + cos θ

1 + cos θ

sec θ

sec θ + 1

Exercises

For Exercises 1-8, prove the given identity.

1. cos 3 θ = 4 cos3 θ − 3 cos θ

2. tan 1 θ

2

= csc θ − cot θ

sin 2 θ

cos 2 θ

sin 3 θ

cos 3 θ

3.

−

= sec θ

4.

−

= 2

sin θ

cos θ

sin θ

cos θ

2

3 tan θ

5. tan 2 θ =

− tan3 θ

6. tan 3 θ

cot θ − tan θ

=

1 − 3 tan2 θ

tan θ − sin θ

cos2 ψ

1

7. tan2 1 θ

+ cos 2 ψ

2

=

8.

tan θ

=

+ sin θ

cos2 θ

1 + cos 2 θ

9. Some trigonometry textbooks used to claim incorrectly that sin θ + cos θ =

1 + sin 2 θ was

an identity. Give an example of a specific angle θ that would make that equation false. Is

sin θ + cos θ = ± 1 + sin 2 θ an identity? Justify your answer.

10. Fill out the rest of the table below for the angles 0◦ < θ < 720◦ in increments of 90◦, showing θ,

1 θ, and the signs (

θ.

2

+ or −) of sin θ and tan 12

θ

1 θ

sin θ

tan 1 θ

θ

1 θ

sin θ

tan 1 θ

2

2

2

2

0◦ − 90◦

0◦ − 45◦

+

+

360◦ − 450◦

180◦ − 225◦

90◦ − 180◦

45◦ − 90◦

450◦ − 540◦

225◦ − 270◦

180◦ − 270◦

90◦ − 135◦

540◦ − 630◦

270◦ − 315◦

270◦ − 360◦

135◦ − 180◦

630◦ − 720◦

315◦ − 360◦

11. In general, what is the largest value that sin θ cos θ can take? Justify your answer.

For Exercises 12-17, prove the given identity for any right triangle △ ABC with C = 90◦.

12. sin ( A − B) = cos 2 B

13. cos ( A − B) = sin 2 A

2 ab

b 2

14. sin 2 A

− a 2

=

c 2

15. cos 2 A =

c 2

2 ab

c − b

a

16. tan 2 A =

17. tan 1 A =

=

b 2 − a 2

2

a

c + b

18. Continuing Exercise 20 from Section 3.1, it can be shown that

r (1 − cos θ) = a (1 + ǫ)(1 − cos ψ) , and

r (1 + cos θ) = a (1 − ǫ)(1 + cos ψ) ,

where θ and ψ are always in the same quadrant. Show that tan 1 θ

tan 1 ψ .

2

=

1 + ǫ

1 − ǫ

2

82

Chapter 3 • Identities

§3.4

3.4 Other Identities

Though the identities in this section fall under the category of “other”, they are perhaps

(along with cos2 θ +sin2 θ = 1) the most widely used identities in practice. It is very common

to encounter terms such as sin A + sin B or sin A cos B in calculations, so we will now

derive identities for those expressions. First, we have what are often called the product-to-

sum formulas:

sin A cos B =

1 (sin ( A

2

+ B) + sin ( A − B))

(3.37)

cos A sin B =

1 (sin ( A

2

+ B) − sin ( A − B))

(3.38)

cos A cos B =

1 (cos ( A

2

+ B) + cos ( A − B))

(3.39)

sin A sin B = − 1 (cos ( A

2

+ B) − cos ( A − B))

(3.40)

We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We

see that

sin ( A + B) + sin ( A − B) = (sin A cos B + ✭✭✭✭✭✭

cos A sin B) + (sin A cos B − ✭✭✭✭✭✭

cos A sin B)

= 2 sin A cos B ,

so formula (3.37) follows upon dividing both sides by 2. Notice how in each of the above

identities a product (e.g. sin A cos B) of trigonometric functions is shown to be equivalent to

a sum (e.g. 1 (sin ( A

2

+ B) + sin ( A− B))) of such functions. We can go in the opposite direction,

with the sum-to-product formulas:

sin A + sin B =

2 sin 1 ( A

( A

2

+ B) cos 12

− B)

(3.41)

sin A − sin B =

2 cos 1 ( A

( A

2

+ B) sin 12

− B)

(3.42)

cos A + cos B =

2 cos 1 ( A

( A

2

+ B) cos 12

− B)

(3.43)

cos A − cos B = −2 sin 1 ( A

( A

2

+ B) sin 12

− B)

(3.44)

These formulas are just the product-to-sum formulas rewritten by using some clever sub-

stitutions: let x = 1 ( A

( A

2

+ B) and y = 12

− B). Then x + y = A and x − y = B. For example, to

derive formula (3.43), make the above substitutions in formula (3.39) to get

cos A + cos B = cos ( x + y) + cos ( x − y)

= 2 · 1 (cos ( x

2

+ y) + cos ( x − y))

= 2 cos x cos y

(by formula (3.39))

= 2 cos 1 ( A

( A

2

+ B) cos 12

− B) .

The proofs of the other sum-to-product formulas are similar (see Exercises 4-6).

Other Identities • Section 3.4

83

Example 3.16

We are now in a position to prove Mollweide’s equations, which we introduced in Section 2.3: For any

triangle △ ABC,

a − b

sin 1 ( A − B)

a + b

cos 1 ( A − B)

=

2

and

=

2

.

c

cos 1 C

c

sin 1 C

2

2

First, since C = 2 · 1 C, by the double-angle formula we have sin C

C cos 1 C. Thus,

2

= 2 sin 12

2

a − b

a

b

sin A

sin B

=

−

=

−

(by the Law of Sines)

c

c

c

sin C

sin C

sin A − sin B

sin A − sin B

=

=

sin C

2 sin 1 C cos 1 C

2

2

2 cos 1 ( A + B) sin 1 ( A − B)

=

2

2

(by formula (3.42))

2 sin 1 C cos 1 C

2

2

cos 1 (180◦ − C) sin 1 ( A − B)

=

2

2

(since A + B = 180◦ − C)

sin 1 C cos 1 C

2

2

✘✘✘✘✘✘

cos (90◦ − 1 C) sin 1 ( A − B)

=

2

2

✟✟✟

sin 1 C cos 1 C

2

2

sin 1 ( A − B)

=

2

(since cos (90◦ − 1 C) = sin 1 C) .

cos 1 C

2

2

2

This proves the first equation. The proof of the other equation is similar (see Exercise 7).

Example 3.17

Using Mollweide’s equations, we can prove the Law of Tangents: For any triangle △ ABC,

a − b

tan 1 ( A − B)

b − c

tan 1 ( B − C)

c − a

tan 1 ( C − A)

=

2

,

=

2

,

=

2

.

a + b

tan 1 ( A

b + c

tan 1 ( B

c + a

tan 1 ( C

2

+ B)

2

+ C)

2

+ A)

We need only prove the first equation; the other two are obtained by cycling through the letters. We

see that

sin 1 ( A − B)

a − b

2

a − b

c

cos 1 C

=

=

2

(by Mollweide’s equations)

a + b

a + b

cos 1 ( A

2

− B)

c

sin 1 C

2

sin 1 ( A − B)

sin 1 C

=

2

·

2

cos 1 ( A

cos 1 C

2

− B)

2

= tan 1 ( A

C

( A

( A

2

− B) · tan 12

= tan 12

− B) · tan (90◦ − 12

+ B)) (since C = 180◦ − ( A + B))

= tan 1 ( A

( A

( A

( A

2

− B) · cot 12

+ B) (since tan (90◦ − 12

+ B)) = cot 12

+ B), see Section 1.5)

tan 1 ( A − B)

=

2

.

QED

tan 1 ( A

2

+ B)

84

Chapter 3 • Identities

§3.4

Example 3.18

For any triangle △ ABC, show that

cos A + cos B + cos C = 1 + 4 sin 1 A sin 1 B sin 1 C .

2

2

2

Solution: Since cos ( A + B + C) = cos 180◦ = −1, we can rewrite the left side as

cos A + cos B + cos C = 1 + (cos ( A + B + C) + cos C) + (cos A + cos B) , so by formula (3.43)

= 1 + 2 cos 1 ( A

( A

( A

( A

2

+ B + 2 C) cos 12

+ B) + 2 cos 12

+ B) cos 12<