4.

4 Double-Measure and Half-Measure Identities

4.4.1. Double – Measure Identities
Double – Measure Identities for Sine

sin 2𝑥 = 2 sin 𝑥 cos 𝑥

Double – Measure Identities for Cosine

cos 2𝑥 = cos 2 𝑥 − sin2 𝑥

Or

cos 2𝑥 = 1 − 2 sin2 𝑥

Or

cos 2𝑥 = 2 cos 2 𝑥 − 1

Double – Measure Identities for Tangent

2 tan 𝑥
tan 2𝑥 =
1 − tan 𝑥

24
Example 1: Given tan 𝜃 = , 𝜃 in Q III. Find:
7
a. sin 2𝜃 e. csc 2𝜃
b. cos 2𝜃 f. sec 2𝜃
c. tan 2𝜃 g. Quadrant where 2𝜃 is located
d. cot 2𝜃
24 y
tan  = =
7 x
x = −7
y = −24
r= x2 + y2
r= (− 7 )2 + (− 24 )2
r = 49 + 576
r = 25

y 24 x 7
sin  = =− cos = =−
r 25 r 25
a. sin 2𝜃 b. cos 2𝜃
Solution: Solution:
sin 2 = 2 sin  cos cos 2 = cos2  − sin 2 
 24  7  2
 7   24 
2
= 2 −  −  = −  − − 
 25  25   25   25 
336
= =
49 576
−
625 625 625
527
=−
625

c. tan 2𝜃 d. cot 2𝜃
Solution: Solution:
2 tan  cot 2 =
1
tan 2 = tan 2
1 − tan 
 24  1
2  =
2 tan 
=  
7
 24  1 − tan 
1−   1
 7  =
 24 
48 2 
 7 
= 7  24 
17 1−  
−
7  7 
 48  7  =
1
=   − 
 7  17  48
48 7
=− 17
17 −
7
1
=
 48  7 
  − 
 7  17 
1
=
48
−
17
 17 
= 1 − 
 48 
17
=−
48
 e. csc 2𝜃 f. sec 2𝜃
Solution: Solution:
1
csc 2 = 1
sin 2 sec 2 =
cos 2
1
= 1
2 sin  cos =
cos  − sin 2 
2
1
= 1
 24  7  =
2 −  −   7 
2
 24 
2

 25  25  −  − − 
 25   25 
1
= 1
336 =
49 576
625 −
625 625
 625 
= 1  1
 336  =
527
625 −
= 625
336  625 
= 1 − 
 527 
625
=−
527

g. Quadrant where 2𝜃 is located.

Ans: Quadrant II

3
Example 2: Given sin 𝜃 = , 𝜃 in Q II. Find sin 3𝜃?
5
Solution:
3 y
sin  = = sin 3 = sin (2 +  )
5 r
= sin 2 cos + cos 2 sin 
y=3
r =5 = (2 sin  cos ) cos + (cos2  − sin 2  )sin 
= 2 sin  cos2  + cos2  sin  − sin 3 
x=− r −y 2 2
2 2 3
 3  4   4  3  3
x=− (5) 2
− (3)
2
= 2  −  +  −    −  
 5  5   5 5 5
x = − 25 − 9 96 48 27
= + −
x = −4 125 125 125
x 4 117
cos = = − =
r 5 125
 4.4.2. Half – Measure Identities

Half – Measure Identities for Sine

𝑥 1 − cos 𝑥
sin = ±√
2 2

Half – Measure Identities for Cosine

𝑥 1 + cos 𝑥
cos = ±√
2 2

Half – Measure Identities for Tangent

𝑥 sin 𝑥 𝑥 1−cos 𝑥
tan = and tan =
2 1+cos 𝑥 2 sin 𝑥

Example 1: Find the exact value of the following.

1. sin 750 2. cos 750
Solution: Solution:
0
150 150 0
sin 75 0 = sin cos 75 0 = cos
2 2
1 − cos150 0 1 + cos150 0
= =
2 2
 3  3
1 −  − 
 1 +  − 

 2   2 
=+ =+
2 2
2+ 3 2− 3
=+ 2 =+ 2
2 2
 2 + 3  1   2 − 3  1 
= +   
 = +   

 2  2   2  2 
2+ 3 2− 3
=+ =+
4 4
2+ 3 2− 3
= =
2 2
 3. tan 750
Solution:
150 0
tan 75 0 = tan
2
sin 150 0
=
1 + cos150 0
1
= 2
 3
1 +  − 

 2 
1
= 2
2− 3
2
 1  2 
=   
 2  2 − 3 
1
=
2− 3
1 2+ 3
=  
2 − 3  2 + 3 
2+ 3
=
(2)2 − ( )
2
3
= 2+ 3

5
Example 2: Given sec 𝛼 = , 𝛼 in Q IV, find the exact values of the following:
3
𝛼
a. csc 𝛼 =
2
𝛼
b. cot 𝛼 =
2
5 r
sec  = =
3 x
x=3
r =5
y = − r 2 − x2
y=− (5)2 − (3)2
y = − 25 − 9
y = −4
y 4 x 3
sin  = = − cos = =
r 5 r 5
 𝛼 𝛼
a. csc b. cot
2 2
Solution: Solution:
 1  1
csc = cot =
2  2 
sin tan
2 2
1 1
= =
 
sin tan
2 2
1 1
= =
1 − cos sin 
2 1 + cos
1 1
= =
3 4
1− −
5 5
3
2 1+
5
1 1
= =
2 4
−
5 5
2 8
5
1 1
= =
 2  1   4  5 
    −  
 5  2   5  8 
1 1
= =
1 1
−
5 2
 
 
 1   2
= = 1 − 
 1   1
 
 5
 5
= 1 
 = −2
 1 
= 5
 Activity 4.4
Double-Measure and Half-Measure Identities
Name: Date:
Year & Section: Score:
Solve the following as indicated and show your complete solution.

7
I. If cos 𝛼 = , 𝛼 𝜖 Q IV, find
10
a. sin 2𝛼
b. cos 2𝛼
c. tan 2𝛼
d. Quadrant where 2𝛼 is located.
7
II. Given 𝛼 = 𝜋, find the exact values of the following:
6
𝛼
a. sin
2
𝛼
b. cos
2
𝛼
c. tan
2