identity is likely to be easier to expand. The reason is that, by its complexity, there will be
more things that you can do with that expression. For example, if you were asked to prove
that
sec θ − sin θ tan θ = cos θ ,
there would not be much that you could do with the right side of that identity; it consists of
a single term (cos θ) that offers no obvious means of expansion.
Example 3.4
1 + cot2 θ
Prove that
= csc θ cot θ .
sec θ
Solution: Of the two sides, the left side looks more complicated, so we will expand that:
1 + cot2 θ
csc2 θ
=
(by (3.11))
sec θ
sec θ
1
csc θ · sin θ
=
1
cos θ
cos θ
= csc θ · sin θ
= csc θ cot θ
(by (3.2))
Basic Trigonometric Identities • Section 3.1
69
Example 3.5
tan2 θ + 2
Prove that
= 1 + cos2 θ .
1 + tan2 θ
Solution: Expand the left side:
tan2 θ + 2
tan2 θ + 1 + 1
=
1 + tan2 θ
1 + tan2 θ
sec2 θ + 1
=
(by (3.10))
sec2 θ
sec2 θ
1
=
+
sec2 θ
sec2 θ
= 1 + cos2 θ
When trying to prove an identity where at least one side is a ratio of expressions, cross-
multiplying can be an effective technique:
a
c
=
if and only if
ad = bc
b
d
Example 3.6
1 + sin θ
cos θ
Prove that
=
.
cos θ
1 − sin θ
Solution: Cross-multiply and reduce both sides until it is clear that they are equal:
(1 + sin θ)(1 − sin θ) = cos θ · cos θ
1 − sin2 θ = cos2 θ
By (3.5) both sides of the last equation are indeed equal. Thus, the original identity holds.
Example 3.7
Suppose that a cos θ = b and c sin θ = d for some angle θ and some constants a, b, c, and d. Show that a 2 c 2 = b 2 c 2 + a 2 d 2.
Solution: Multiply both sides of the first equation by c and the second equation by a:
ac cos θ = bc
ac sin θ = ad
Now square each of the above equations then add them together to get:
( ac cos θ)2 + ( ac sin θ)2 = ( bc)2 + ( ad)2
( ac)2 cos2 θ + sin2 θ = b 2 c 2 + a 2 d 2
a 2 c 2 = b 2 c 2 + a 2 d 2
(by (3.3))
Notice how θ does not appear in our final result. The trick was to get a common coefficient ( ac) for
cos θ and sin θ so that we could use cos2 θ + sin2 θ = 1. This is a common technique for eliminating
trigonometric functions from systems of equations.
70
Chapter 3 • Identities
§3.1
Exercises
1. We showed that sin θ = ± 1 − cos2 θ for all θ. Give an example of an angle θ such that
sin θ = − 1 − cos2 θ .
2. We showed that cos θ = ± 1 − sin2 θ for all θ. Give an example of an angle θ such that
cos θ = − 1 − sin2 θ .
3. Suppose that you are given a system of two equations of the following form:1
A cos φ = B ν 1 − Bν 2 cos θ
A sin φ = B ν 2 sin θ .
Show that A 2 = B 2 ν 21 + ν 22 − 2 ν 1 ν 2 cos θ .
For Exercises 4-16, prove the given identity.
4. cos θ tan θ = sin θ
5. sin θ cot θ = cos θ
tan θ
csc θ
6.
= tan2 θ
7.
= csc2 θ
cot θ
sin θ
cos2 θ
1 − 2 cos2 θ
8.
= 1 − sin θ
9.
= tan θ − cot θ
1 + sin θ
sin θ cos θ
10. sin4 θ − cos4 θ = sin2 θ − cos2 θ
11. cos4 θ − sin4 θ = 1 − 2 sin2 θ
1 − tan θ
cot θ − 1
tan θ + tan φ
12.
=
13.
= tan θ tan φ
1 + tan θ
cot θ + 1
cot θ + cot φ
sin2 θ
1 − tan2 θ
14.
= tan2 θ
15.
= 1 − sec2 θ
1 − sin2 θ
1 − cot2 θ
tan θ
16. sin θ = ±
( Hint: Solve for sin2 θ in Exercise 14. )
1 + tan2 θ
17. Sometimes identities can be proved by geometrical methods. For ex-
y
(1, y)
ample, to prove the identity in Exercise 16, draw an acute angle θ in QI
y
and pick the point (1, y) on its terminal side, as in Figure 3.1.2. What
θ
must y equal? Use that to prove the identity for acute θ. Explain the
x
0
adjustment(s) you would need to make in Figure 3.1.2 to prove the iden-
1
tity for θ in the other quadrants. Does the identity hold if θ is on either
axis?
Figure 3.1.2
18. Similar to Exercise 16 , find an expression for cos θ solely in terms of tan θ.
19. Find an expression for tan θ solely in terms of sin θ, and one solely in terms of cos θ.
20. Suppose that a point with coordinates ( x, y) = ( a (cos ψ − ǫ), a 1 − ǫ 2 sin ψ) is a distance r > 0
from the origin, where a > 0 and 0 < ǫ < 1. Use r 2 = x 2 + y 2 to show that r = a (1 − ǫ cos ψ) .
(Note: These coordinates arise in the study of elliptical orbits of planets.)
21. Show that each trigonometric function can be put in terms of the sine function.
1These types of equations arise in physics, e.g. in the study of photon-electron collisions. See pp. 95-97 in W.
RINDLER, Special Relativity, Edinburgh: Oliver and Boyd, LTD., 1960.
Sum and Difference Formulas • Section 3.2
71
3.2 Sum and Difference Formulas
We will now derive identities for the trigonometric functions of the sum and difference of
two angles. For the sum of any two angles A and B, we have the addition formulas:
sin ( A + B) = sin A cos B + cos A sin B
(3.12)
cos ( A + B) = cos A cos B − sin A sin B
(3.13)
To prove these, first assume that A and B are acute angles. Then A + B is either acute or
obtuse, as in Figure 3.2.1. Note in both cases that ∠ QPR = A, since
∠ QPR = ∠ QPO − ∠ OP M = (90◦ − B) − (90◦ − ( A + B)) = A in Figure 3.2.1(a), and
∠ QPR = ∠ QPO + ∠ OP M = (90◦ − B) + (90◦ − (180◦ − ( A + B))) = A in Figure 3.2.1(b).
P
P
A
A
Q
R
A + B
R
Q
A + B
B
B
A
A
O
N
N
M
M
O
(a) A + B acute
(b) A + B obtuse
Figure 3.2.1
sin ( A + B) and cos ( A + B) for acute A and B
Thus,
MP
MR + RP
NQ + RP
NQ
RP
sin ( A + B) =
=
=
=
+
OP
OP
OP
OP
OP
NQ
OQ
RP
PQ
=
·
+
·
OQ
OP
PQ
OP
= sin A cos B + cos A sin B ,
(3.14)
72
Chapter 3 • Identities
§3.2
and
OM
ON − MN
ON − RQ
ON
RQ
cos ( A + B) =
=
=
=
−
OP
OP
OP
OP
OP
ON
OQ
RQ
PQ
=
·
+
·
OQ
OP
PQ
OP
= cos A cos B − sin A sin B .
(3.15)
So we have proved the identities for acute angles A and B. It is simple to verify that they
hold in the special case of A = B = 0◦. For general angles, we will need to use the relations
we derived in Section 1.5 which involve adding or subtracting 90◦:
sin ( θ + 90◦) =
cos θ
sin ( θ − 90◦) = −cos θ
cos ( θ + 90◦) = −sin θ
cos ( θ − 90◦) =
sin θ
These will be useful because any angle can be written as the sum of an acute angle (or
0◦ ) and integer multiples of ±90◦. For example, 155◦ = 65◦ + 90◦, 222◦ = 42◦ + 2(90◦), −77◦ =
13◦ − 90◦, etc. So if we can prove that the identities hold when adding or subtracting 90◦
to or from either A or B, respectively, where A and B are acute or 0◦, then the identities
will also hold when repeatedly adding or subtracting 90◦, and hence will hold for all angles.
Replacing A by A + 90◦ and using the relations for adding 90◦ gives
sin (( A + 90◦) + B) = sin (( A + B) + 90◦) = cos ( A + B) ,
= cos A cos B − sin A sin B (by equation (3.15))
= sin ( A + 90◦) cos B + cos ( A + 90◦) sin B ,
so the identity holds for A + 90◦ and B (and, similarly, for A and B + 90◦). Likewise,
sin (( A − 90◦) + B) = sin (( A + B) − 90◦) = −cos ( A + B) ,
= −(cos A cos B − sin A sin B)
= (−cos A) cos B + sin A sin B
= sin ( A − 90◦) cos B + cos ( A − 90◦) sin B ,
so the identity holds for A −90◦ and B (and, similarly, for A and B +90◦). Thus, the addition
formula (3.12) for sine holds for all A and B. A similar argument shows that the addition
formula (3.13) for cosine is true for all A and B.
QED
Replacing B by − B in the addition formulas and using the relations sin (− θ) = −sin θ and
cos (− θ) = cos θ from Section 1.5 gives us the subtraction formulas:
sin ( A − B) = sin A cos B − cos A sin B
(3.16)
cos ( A − B) = cos A cos B + sin A sin B
(3.17)
Sum and Difference Formulas • Section 3.2
73
Using the identity tan θ = sin θ , and the addition formulas for sine and cosine, we can
cos θ
derive the addition formula for tangent:
sin ( A + B)
tan ( A + B) = cos ( A + B)
sin A cos B + cos A sin B
= cos A cos B − sin A sin B
sin A cos B
cos A sin B
+
cos A cos B
cos A cos B
=
(divide top and bottom by cos A cos B)
cos A cos B
sin A sin B
−
cos A cos B
cos A cos B
sin A
cos B
cos A
sin B
·
+
·
cos A
cos B
cos A
cos B
tan A + tan B
=
=
sin A
sin B
1 − tan A tan B
1 −
·
cos A
cos B
This, combined with replacing B by − B and using the relation tan (− θ) = −tan θ, gives us
the addition and subtraction formulas for tangent:
tan A + tan B
tan ( A + B) =
(3.18)
1 − tan A tan B
tan A − tan B
tan ( A − B) =
(3.19)
1 + tan A tan B
Example 3.8
Given angles A and B such that sin A = 4 , cos A
, sin B
, and cos B
, find the exact values
5
= 35
= 12
13
= 5
13
of sin ( A + B), cos ( A + B), and tan ( A + B).
Solution: Using the addition formula for sine, we get:
sin ( A + B) = sin A cos B + cos A sin B
4
5
3
12
56
=
·
+
·
⇒
sin ( A + B) =
5
13
5
13
65
Using the addition formula for cosine, we get:
cos ( A + B) = cos A cos B − sin