Pre-calculus

1. STEM PRE – CALCULUS FEU Senior High School
1 | P a g e
LESSON 7 – DOUBLE-ANGLE AND HALF-ANGLE IDENTITIES
INTRODUCTION
Trigonometric identities simplify the computations of trigonometric expressions. In this
lesson, we continue establishing more trigonometric identities. We derive the formulas for f (2𝜃)
and 𝑓(1/2𝜃), where 𝑓 is the sine, cosine, or tangent function.
OBJECTIVES
At the end of the lesson, you are expected to:
a) derive the double-angle and half-angle identities;
b) simplify trigonometric expressions using known identities;
c) prove other trigonometric identities using known identities; and
d) solve situational problems involving trigonometric identities.
SUBJECT MATTER
In this lesson you will use formulas for double angles (angles of measure 2𝑢) and half
angles (angles of measure
𝑢
2
). The three formulas for cos 2𝑢 below are equivalent, as are the two
formulas for 𝑡𝑎𝑛
𝑢
2
. Use whichever formula is most convenient for solving a problem.
Double-Angle Identities
Recall the sum identities for sine and cosine.
sin( 𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵
cos( 𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
When 𝐴 = 𝐵, these identities becomes
sin 2𝐴 = sin 𝐴 cos 𝐴 + cos 𝐴 sin 𝐴 = 2 sin 𝐴 cos 𝐴
and
cos 2𝐴 = cos 𝐴 cos 𝐴 − sin 𝐴 sin 𝐴 = cos2
𝐴 − sin2
𝐴
The double-angle identity for cosine has other forms. We use the Pythagorean identity
sin2
𝜃 + cos2
𝜃 = 1.
cos 2𝐴 = cos2
𝐴−sin2
𝐴
= cos2
𝐴 − (1 − cos2
𝐴)
= 2 cos2
𝐴 − 1
cos 2𝐴 = cos2
𝐴−sin2
𝐴
= (1 − sin2
𝐴) − sin2
𝐴
= 1 − 2sin2
𝐴

2. STEM PRE – CALCULUS FEU Senior High School
2 | P a g e
Example 3.7.1: Verify ((sin 𝐴 + cos 𝐴)2
= 1 + sin 2𝐴
Solution:
Example 3.7.2: Find the exact value of cos 2𝐴 if sin 𝐴 = −
12
13
in Quadrant III.
Solution:Using the following form of the cosine of a double angle formula,
𝑐𝑜𝑠 2𝐴 = 1 − 2 sin2
𝐴, we have:
Example 3.7.3: Given sin 𝑡 =
3
5
and
𝜋
2
< 𝑡 < 𝜋, find sin 2𝑡 and cos 2𝑡.
Solution: We first find cos 𝑡 using the Pythagorean identity. Since 𝑡 lies in QII, we have
cos 𝑡 = −√1 − sin2 𝑡 = −√1 − (
3
5
)
2
= −
4
5
sin 2𝑡 = 2 sin 𝑡 cos 𝑡 cos 2𝑡 = 1 − 2sin2
𝑡
= 2 (
3
5
)(−
4
5
) = 1 − 2 (
3
5
)
2
In the last example, we may compute cos 2𝑡 using one of the other two double-angle
identities for cosine. For the sake of answering the curious minds, we include the computation
here.
In the three cosine double-angle identities, which formula to use depends on the
convenience, what is given, and what is asked.
(sin 𝐴 + cos 𝐴)(sin 𝐴 + cos 𝐴) = 1 + sin 2𝐴
sin2
𝐴 + sin 𝐴 cos 𝐴 + sin 𝐴 cos 𝐴 + cos2
𝐴 = 1 + sin 2𝐴
sin2
𝐴 + 2sin𝐴cos𝐴 + cos2
𝐴 = 1 + sin 2𝐴
sin2
𝐴 + cos2
𝐴 + sin 2𝐴 = 1 + sin 2𝐴
1 + sin 2𝐴 = 1 + sin 2𝐴
𝑐𝑜𝑠 2𝐴 = 1 − 2 sin2
𝐴
= 1 − 2 (
−12
13
)
2
= 1 − 2 (
144
169
)
2
=
169 − 288
169
𝑐𝑜𝑠 2𝐴 = −
119
169
= −
24
25
=
7
25
cos 2𝑡 = cos2
𝑡 − sin2
𝑡
= (−
4
5
)
2
− (
3
5
)
2
=
7
25
cos 2𝑡 = cos2
𝑡 − sin2
𝑡
= 1 − 2 (
3
5
)
2
=
7
25

3. STEM PRE – CALCULUS FEU Senior High School
3 | P a g e
Example 3.7.4: If 𝑡𝑎𝑛 𝜃 = −
1
3
and 𝑠𝑒𝑐 𝜃 > 0, find 𝑠𝑖𝑛 2𝜃, 𝑐𝑜𝑠 2𝜃 and 𝑡𝑎𝑛 2𝜃.
Solution: We can compute immediately tan 2𝜃
From the given information, we deduce that 𝜃lies in QIV. Using one Pythagorean identity,
we compute cos 𝜃 through sec 𝜃. We may also use the technique discussed in previous lesson by
solving 𝑥, 𝑦, and 𝑟. Then we proceed to find cos 2𝜃.
Find sin 2𝜃; when cos 2𝜃 =
4
5
and tan 2𝜃 = −
3
4
tan 2𝜃 =
2 tan 𝜃
1 − tan2 𝜃
=
2 (−
1
3
)
1 − (−
1
3
)
2 = −
3
4
sec 𝜃 = √1 + tan2 𝜃 = √1 + (−
1
3
)
2
=
√10
3
cos 𝜃 =
1
sec 𝜃
=
1
√10
3
=
3√10
10
cos 2𝜃 = 2 cos2
𝜃 − 1 = 2 (
3√10
10
)
2
− 1 =
4
5
cos 2𝜃 = 2 cos2
𝜃 − 1 = 2 (
3√10
10
)
2
− 1 =
4
5
cos 2𝜃 =
4
5
tan 2𝜃 = −
3
4
tan 2𝜃 =
sin 2𝜃
cos 2𝜃
sin 2𝜃 = tan 2𝜃 cos 2𝜃
sin 2𝜃 = (−
3
4
) (
4
5
)
sin 2𝜃 = −
3
5

4. STEM PRE – CALCULUS FEU Senior High School
4 | P a g e
Half-Angle Identities
Recall two of the three double-angle identities for cosine:
cos 2𝐴 = 2 cos2
𝐴 − 1 and cos 2𝐴 = 1 − 2 sin2
𝐴
From these identities, we obtain two useful identities expressing sin2
𝐴 and cos2
𝐴 in terms of
cos 2 𝐴 as follows:
cos2
𝐴 =
1 + cos 2𝐴
2
and sin2
𝐴 =
1 − cos 2𝐴
2
From these identities, replacing 𝐴 with
𝐴
2
, we get
cos2
𝐴
2
=
1 + cos 2 (
𝐴
2
)
2
=
1 + cos 𝐴
2
and
sin2
𝐴
2
=
1 − cos 2 (
𝐴
2
)
2
=
1 − cos 𝐴
2
These are the half-angle identities for sine and cosine.
Because of the “square” in the formulas, we get
The appropriate signs of cos
𝐴
2
and sin
𝐴
2
depend on which quadrant
𝐴
2
lies.
cos
𝐴
2
= ±√
1 + cos 𝐴
2
and sin
𝐴
2
= ±√
1 − cos 𝐴
2
tan
𝐴
2
=
1 − cos 𝐴
sin 𝐴
and tan
𝐴
2
=
sin 𝐴
1 + cos 𝐴

5. STEM PRE – CALCULUS FEU Senior High School
5 | P a g e
Example 3.7.5: Find the exact value of tan
𝜋
8
.
Solution: Use the fact the
𝜋
8
is half of
𝜋
4
, then
tan
𝜋
8
= tan
1
2
(
𝜋
4
)
Example 3.7.6: Find the exact value of cos 105°.
Solution: Use the fact that 105°.is half of 210° and that cosine is negative in Quadrant III
cos 105° = cos
1
2
(210°)
cos 105° = −√
1 + cos 210°
2
= −
√
1 + (−
√3
2
)
2
= −√
2 − √3
4
= −
√2 − √3
2
Example 3.7.7: Find the exact values of sin 22.5º and cos 22.5º
Solution: Clearly, 22.5° lies in QI (and so sin 22.5º and cos 22.5º are both positive), and 22.5º is
the half-angle of 45º.
sin 22.5° = √
1 − cos 45°
2
=
√1 −
√2
2
2
= √
2 − √2
2
cos 22.5° = √
1 + cos 45°
2
=
√1 +
√2
2
2
= √
2 + √2
2
tan
𝜋
8
=
1 − 𝑐𝑜𝑠
𝜋
4
𝑠𝑖𝑛
𝜋
4
=
1 −
√2
2
√2
2
=
2 − √2
√2
= √2 − 1

6. STEM PRE – CALCULUS FEU Senior High School
6 | P a g e
Example 3.7.8: Find the exact value of tan
𝜋
12
.
Solution:
tan
𝜋
12
=
1 − cos
𝜋
6
sin
𝜋
6
=
1 −
√3
2
1
2
= 2 − √3
EVALUATION
1. Express tan 3𝜃 in terms of tan𝜃
2. Given 𝑐𝑜𝑠 𝑢 = −
3
5
with 𝜋 < 𝑢 <
3𝜋
2
, find 𝑠𝑖𝑛 2𝑢.
3. Given 𝑐𝑜𝑠 𝑢 = −
3
5
with 𝜋 < 𝑢 <
3𝜋
2
, find 𝑠𝑖𝑛
𝑢
2
.
4.
𝒄𝒐𝒔 𝟐𝜽
𝒔𝒊𝒏 𝜽+𝒄𝒐𝒔 𝜽
5. 𝑠𝑖𝑛 3𝑥 = 3𝑠𝑖𝑛𝑥 − 4𝑠𝑖𝑛3
𝑥
HOMEWORK