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<