Trigonometric
Identities
4.3
Introduction
A trigonometric identity is a relation between trigonometric expressions which is true for all
values of the variables (usually angles). There are a very large number of such identities. In this
Section we discuss only the most important and widely used. Any engineer using trigonometry
in an application is likely to encounter some of these identities.
Prerequisites
① have a basic knowledge of the geometry of
triangles
Before starting this Section you should . . .
Learning Outcomes
✓ be familiar with the main trigonometric
identities
After completing this Section you should be
able to . . .
✓ use trigonometric identities to combine
trigonometric functions
1. Trigonometric Identities
A trigonometric identity is simply a relation which is always true.
1
etc., evaluate sin2 θ + cos2 θ for
2
(iii) θ = 60◦ .
Using the familiar values sin 30◦ =
(i) θ = 30◦
(ii) θ = 45◦
[Note that sin2 θ ≡ (sin θ)2 , cos2 θ ≡ (cos θ)2 ]
Your solution
2 √ 2
3
1
1 3
=
+
= + =1
2
2
4 4
2 2
1
1
1 1
√
=
+ √
= + =1
2 2
2
2
√ 2 2
1
3
+
=1
2
2
sin2 60◦ + cos2 60◦ =
sin2 45◦ + cos2 45◦
sin2 30◦ + cos2 30◦
Key Point
For any value of θ
sin2 θ + cos2 θ = 1
(5)
One way of demonstrating this result is to use the definitions of sin θ and cos θ obtained from
the circle of unit radius. Refer back to Figure 10.
Recall that cos θ = OQ, sin θ = OR = P Q but, by Pythagoras’ Theorem
(OQ)2 + (QP )2 = (OP )2 = 1
hence cos2 θ + sin2 θ = 1.
We have demonstrated the result (5) using an angle θ in the first quadrant but the result is true
for any θ i.e. it is indeed an identity.
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
2
By dividing the identity
sin2 θ + cos2 θ = 1
by (i) sin2 θ (ii) cos2 θ obtain further identities.
Hint: Recall the definitions of cosec θ, sec θ, cot θ.
Your solution
tan2 θ + 1 = sec2 θ
1
sin2 θ cos2 θ
+
=
cos2 θ cos2 θ
cos2 θ
(ii)
1 + cot2 θ = cosec2 θ
1
sin2 θ cos2 θ
+
=
sin2 θ sin2 θ
sin2 θ
(i)
We now prove a trigonometric identity from which most of the other important identities can
be easily obtained. We shall adopt the notation for a triangle of denoting by a the length of the
side opposite angle A, b the length of side opposite angle B and c that side opposite angle C.
Refer to Figure 19 where we have shown a quarter of a circle of unit radius.
1
1
P2
P2
b
d
P1
a
B
A
b
P1
a
B
c
A
A
O
1
O
(a)
Q
R
1
(b)
Figure 19
Clearly in the right-angled triangle in Figure 19 (a) we have
sin B =
3
b
=b
OP2
cos B =
a
=a
OP2
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
If the sum of angles A and B is denoted by C we have, from Figure 19(b)
sin C = P2 Q = c + d
= a sin A + b cos A
Substituting for a and b gives
sin(A + B) = sin A cos B + sin B cos A.
Carry out a similar exercise using the same diagrams to obtain the corresponding formula for cos(A + B).
Your solution
cos C = OQ = OR − QR
= a cos A − b sin A
= cos B cos A − sin B sin A
We have, from Figure 19(b)
The following are key identities
Key Point
sin(A + B) = sin A cos B + cos A sin B
(6)
cos(A + B) = cos A cos B − sin A sin B
(7)
Note carefully the addition sign in (6) but the subtraction sign in (7).
Further identities can readily be obtained from (6) and (7).
Replacing B by −B and remembering that cos(−B) = cos B, sin(−B) = − sin B we find
sin(A − B) = sin A cos B − cos A sin B
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
(8)
4
cos(A − B) = cos A cos B + sin A sin B
(9)
Dividing (6) by (7) we obtain
tan(A + B) =
sin(A + B)
cos(A + B)
sin A cos B + cos A sin B
cos A cos B − sin A sin B
Dividing every term by cos A cos B we obtain
tan A + tan B
tan(A + B) =
1 − tan A tan B
=
(10)
By dividing (8) by (9) obtain a similar expansion of tan(A − B)
Your solution
tan(A − B) =
tan A − tan B
1 + tan A tan B
(11)
Dividing every term by cos A cos B gives
tan(A − B) =
sin A cos B − cos A sin B
cos A cos B + sin A sin B
Example Obtain expressions for cos θ in terms of the sine function and for sin θ in terms
of the cosine function.
Solution
Using (9) with A = θ, B = π2 we obtain
cos θ − π2 = cos θ cos π2 + sin θ sin π2 = (cos θ) (0) + sin θ (1)
i.e. sin θ = cos θ − π2 = cos π2 − θ
This result explains why the graph of sin θ has exactly the same shape as the graph of cos θ
π
but it is shifted to the right
by 2 . (See Figure 17). A similar calculation using (6) yields the
π
result cos θ = sin θ + 2 .
5
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
Double angle formulae
If we put B = A in the identity given in (6) we obtain
sin 2A = sin A cos A + cos A sin A
= 2 sin A cos A
(12)
Put B = A in (7) to obtain an identity for cos 2A. Using sin2 A + cos2 A = 1
obtain two alternative forms of the identity.
Your solution
cos 2A = (1 − sin2 A) − sin2 A
cos 2A = 1 − 2 sin2 A
(15)
Alternatively substituting for cos2 A in (13)
(14)
cos 2A = cos2 A − (1 − cos2 A)
= 2 cos2 A − 1
Substituting for sin2 A in (13) we obtain
. ..
(13)
cos(2A) = (cos A)(cos A) − (sin A)(sin A)
cos(2A) = cos2 A − sin2 A
Using (7) with A = B
Use (14) and (15) to obtain, respectively, cos2 A and sin2 A in terms of cos 2A
Your solution
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
6
1
From (14) cos2 A = (1 + cos 2A)
2
1
From (15) sin2 A = (1 − cos 2A)
2
Use (12) and (13) to obtain an identity for tan 2A in terms of tan A.
Your solution
tan 2A =
sin A
2 cos
A
sin2 A
cos2 A
1−
=
2 tan A
1 − tan2 A
(16)
Dividing numerator and denominator by cos2 A we obtain
tan 2A =
sin 2A
2 sin A cos A
=
cos 2A
cos2 A − sin2 A
Half-angle formula
If we simply replace A by
A
and, consequently 2A by A, in (12) we obtain
2
A
A
sin A = 2 sin
cos
2
2
(17)
Similarly from (13)
A
− 1.
cos A = 2 cos
2
2
(18)
These are examples of half-angle formulae. We can obtain a half-angle formula for tan A using
(16). Replacing A by A2 , 2A by A in (16) we obtain
2 tan A2
tan A =
1 − tan2 A2
7
(19)
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
Other formulae, useful for integration when
functions are present, can be obtained
trigonometric
A
then from (19)
using (19). For example if we let t = tan
2
tan A =
2t
1 − t2
Also, from (17)
A
A
cos
sin A = 2 sin
2
2
A
A
A
2
cos
(multiplying top and bottom by cos
)
= 2 tan
2
2
2
2 tan A2
tan A2
2t
=
= 2 2 A =
2 A
1 + t2
sec 2
1 + tan 2
Use (18) to obtain cos A in terms of t = tan
A
tan2 θ + 1 = sec2 θ with θ =
2
A
2
. Hint: use the identity
Your solution
A
A
A
cos A = 2 cos2
− 1 = cos2
− sin2
2
2
2
A
A
= cos2
1 − tan2
2
2
1 − tan2 A2
1 − tan2 A2
=
=
sec2 A2
1 + tan2 A2
1 − t2
1 + t2
So
cos A =
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
8
Key Point
A
If t = tan
then
2
sin A =
2t
1 + t2
cos A =
1 − t2
1 + t2
tan A =
2t
1 − t2
Sum of two sines and of two cosines
Finally, in this Section , we obtain results that are widely used in areas of science and engineering
such as vibration theory, wave theory and electric circuit theory.
We return to the identities (6) and (8)
sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
Adding these identities gives
sin(A + B) + sin(A − B) = 2 sin A cos B
(20)
Subtracting the identities produces
sin(A + B) − sin(A − B) = 2 cos A sin B
D = A − B so that
It is now convenient to let C = A + B
C −D
C +D
B=
A=
2
2
Hence (20) becomes
sin C + sin D = 2 sin
(21)
C +D
2
cos
C −D
2
(22)
(or, in words, the sum of 2 sines = twice the sine of half the sum (of the angles) multiplied the
cosine of half the difference (of the angles).
Similarly (21) becomes
C −D
C +D
sin
(23)
sin C − sin D = 2 cos
2
2
[Note that (23) can alternatively be obtained by replacing D by −D in (22).]
9
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
Use (7) and (9) in a similar way to the above to obtain results for the sum and
for the difference of two cosines.
Your solution
= 2 sin
C +D
2
cos C − cos D = −2 sin
cos C + cos D = 2 cos
Hence with C = A + B
. ..
D =A−B
C +D
2
C +D
2
cos(A + B)
cos(A − B)
cos(A + B) + cos(A − B)
cos(A + B) − cos(A − B)
sin
=
=
=
=
sin
cos
D−C
2
(25)
C −D
2
C −D
2
(24)
cos A cos B − sin A sin B
cos A cos B + sin A sin B
2 cos A cos B
−2 sin A sin B
Summary
We have covered in this Section a large number of trigonometric identities. The most important
of them and probably the ones most worth memorising are given in the following key-point.
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
10
Key Point
cos2 θ + sin2 θ = 1
sin 2θ = 2 sin θ cos θ
cos 2θ = cos2 θ − sin2 θ
= 2 cos2 θ − 1
= 1 − 2sin2 θ C −D
C +D
cos
sin C + sin D = 2 sin
2
2
C +D
C −D
sin C − sin D = 2 cos
sin
2
2
C +D
C −D
cos C − cos D = −2 sin
sin
2
2
C +D
C −D
cos C + cos D = 2 cos
cos
2
2
11
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
Exercises
1. Show that
sin t sec t = tan t.
2. Show that
(1 + sin t)(1 + sin(−t)) = cos2 t.
3. Show that
1
= 12 sin 2θ.
tan θ + cot θ
4. Show that
sin2 (A + B) − sin2 (A − B) = sin 2A sin 2B.
(Hint: the left hand side is the difference of the squared quantities.)
5. Show that
sin 4θ + sin 2θ
= tan 3θ.
cos 4θ + cos 2θ
6. Show that
cos4 A − sin4 A = cos 2A
7. Express each of the following as the sum (or difference) of 2 sines (or cosines)
(a) sin 5x cos 2x (b) 8 cos 6x cos 4x (c)
1
3
1
sin x cos x
3
2
2
8. Express (a) sin 3θ in terms of cos θ. (b) cos 3θ in terms of cos θ.
9. By writing cos 4x as cos 2(2x), or otherwise, express cos 4x in terms of cos x.
2 tan t
.
2 − sec2 t
10. Show that
tan 2t =
11. Show that
cos 10t − cos 12t
= tan t.
sin 10t + sin 12t
12. Show that the area of an isosceles triangle with equal sides of length x is
A=
x2
sin θ
2
where θ is the angle between the two equal sides. Hint: Use the following figure:
A
θ
2
x
x
B
D
C
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
12
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
1. sin t. sec t = sin t.
13
1
sin t
=
= tan t.
cos t
cos t
2. Recall that sin(−t) = − sin t. (See the graph of sin t if you are unfamiliar with this result.)
Hence
(1 + sin t)(1 + sin(−t)) = (1 + sin t)(1 − sin t)
= 1 − sin2 t
= cos2 t
3.
1
=
tan θ + cos θ
=
sin θ
cos θ
1
+
cos θ
sin θ
=
1
sin2 θ+cos2 θ
sin θ cos θ
sin θ cos θ
sin2 θ + cos2 θ
= sin θ cos θ
=
1
sin 2θ
2
4. Using the hint and the identity
x2 − y 2 = (x − y)(x + y)
we have
sin2 (A + B) − sin2 (A − B) = (sin(A + B) − sin(A − B))(sin(A + B) + sin(A − B))
The first bracket gives
sin A cos B + cos A sin B − (sin A cos B − cos A sin B) = 2 cos A sin B
Similarly the second bracket gives
2 sin A cos B.
Multiplying we obtain (2 cos A sin A)(2 cos B sin B) = sin 2A. sin 2B
5.
sin 4θ + sin 2θ
2 sin 3θ cos θ
sin 3θ
=
=
= tan 3θ
cos 4θ + cos 2θ
2 cos 3θ cos θ
cos 3θ
HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
14
6.
cos4 A − sin4 A =
=
=
=
(cos A)4 − (sin A)4
(cos2 A)2 − (sin2 A)2
(cos2 A − sin2 A)(cos2 A + sin2 A)
cos2 A − sin2 A = cos 2A
7.
(a) Using sin A + sin B = 2 sin
Clearly here
. ..
A+B
2
cos
A−B
2
A−B
= 2x giving A = 7x B = 3x
2
A+B
= 5x
2
1
sin 5x cos 2x = (sin 7x + sin 3x)
2
(b) Using cos A + cos B = 2 cos
A−B
= 4x
2
(c)
A+B
2
cos
A−B
. With
2
A+B
2
= 6x
giving A = 10x B = 2x
. .. 8 cos 6x cos 4x = 4(cos 6x + cos 2x)
1
3x
1
x cos
= (sin 2x − sin x)
2
2
6
1
sin
3
8.
(a)
sin 3θ =
=
=
=
sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ
2 sin θ cos2 θ + (cos2 θ − sin2 θ) sin θ
3 sin θ cos2 θ − sin3 θ
3 sin θ(1 − sin2 θ) − sin3 θ = 3 sin θ − 4 sin3 θ
(b)
cos 3θ =
=
=
=
cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ
(cos2 θ − sin2 θ) cos θ − 2 sin θ cos θ sin θ
cos3 θ − 3 sin2 θ cos θ
cos3 θ − 3(1 − cos2 θ) cos θ = 4 cos3 θ − 3 cos θ
9.
cos 4x = cos 2(2x) =
=
=
=
2 cos2 (2x) − 1
2(cos 2x)2 − 1
2(2 cos2 x − 1)2 − 1
2(4 cos4 x − 4 cos2 x + 1) − 1 = 8 cos4 x − 8 cos2 x + 1.
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HELM (VERSION 1: March 18, 2004): Workbook Level 1
4.3: Trigonometric Identities
10.
tan 2t =
=
11. cos 10t − cos 12t = 2 sin 11t sin t
. ..
2 tan t
1 − tan2 t
2 tan t
2 tan t
=
1 − (sec2 t − 1)
2 − sec2 t
sin 10t + sin 12t = 2 sin 11t cos(−t)
cos 10t − cos 12t
sin t
sin t
=
=
= tan t
sin 10t + sin 12t
cos(−t)
cos t
12. The right-angled triangle ACD has area
CD
θ
=
But
sin
2
x
cos
area of
. ..
area of
. ..
θ
2
=
AD
x
∆ACD =
1
(CD)(AD)
2
θ
. ..
CD = x sin
2
. ..
1 2
x sin
2
AD = x cos
θ
2
cos
θ
2
θ
2
1
= x2 sin θ
4
1
∆ABC = 2 × area of ∆ACD = x2 sin θ
2