Half-angle formulae
Trigonometry
Half-angle formula: sine
We start with the formula for the cosine of a double
angle:
cos 2θ = 1− 2sin2 θ
Now, if we let θ=α/2, then 2θ = α and our formula
becomes:
cos α = 1 − 2sin2 (α/2)
Solving for sin(α/2)
We now solve for sin(α/2) (that is, we get
sin(α/2) on the left of the equation and
everything else on the right):
2sin2(α/2) = 1 − cos α
sin2(α/2) = (1 − cos α)/2
Taking the square root gives us the following
sine of a half-angle identity:

1  cos 
sin  
2
2
Meaning of the ± sign
The sign of depends on the quadrant in which
α/2 lies.
If α/2 is in the first or second quadrants, the
formula uses the positive case:

1  cos 
sin  
2
2
If α/2 is in the third or fourth quadrants, the
formula uses the negative case:

1  cos 
sin  
2
2
Half-angle formulae: cosine
Using a similar process, with the same substitution of
θ=α/2(so 2θ = α) we substitute into the identity
cos 2θ = 2cos2 θ − 1
and we obtain:
cos α = 2 cos2(α/2) – 1
Reverse the equation:
2 cos2(α/2) – 1 = cos α
Half-angle cosine (continued)
Add 1 to both sides:
2 cos2(α/2) = cos α +1
Divide both sides by 2:
cos2(α/2) = (cos α +1)/2
Solving for cos(α/2), we obtain:

1  cos 
cos  
2
2
Meaning of the ± sign (cosine)
As before, the sign we need depends on the quadrant.
• If α/2 is in the first or fourth quadrants, the formula
uses the positive case:

1  cos 
cos  
2
2
• If α/2 is in the second or third quadrants, the
formula uses the negative case:

1  cos 
cos  
2
2
Half-angle formulae: tan
We can develop formulae for tan α/2 by using what we
already know as trigonometry identities.

sin
2
tan 

2
cos 
2

1  cos 
2
1  cos 
2

1  cos 
1  cos 
Half-angle tan (continued calculations)
We now try to get perfect squares inside the square
root:
(1  cos  ) 2

(1  cos  )(1  cos  )
(1  cos  )
1  cos 2 
2

(1  cos  ) 2
1  cos 

2
sin 
sin 
Sign of tangent?
Same considerations of quadrant/s have to be made as
for sine and cosine
Equivalence of tan α/2
Another form:
1  cos  (1  cos  )(1  cos  )

sin 
sin  (1  cos  )
1  cos 2 
sin 2 


sin  (1  cos  ) sin  (1  cos  )
sin 

1  cos 
Summary of tan of half-angles
tan

2
1  cos 

sin 
or
tan

2
sin 

1  cos 
Examples
Example 1: Find the value of sin 15°.
1  cos 
1  cos30
sin  

2
2

2
1  0.866

 0.2588
2
Choose the positive values because 15° is in the 1st
quadrant.
Example 2
Find the value of cos 165° using the cosine half-angle
relationship.
Answer: select α=330° so that α/2=165°.
1  cos 
1  cos 330
cos165  

2
2
1  0.866

 0.9659
2
Minus sign because 165° is in the second quadrant (II).
Example 3
x
Show that 2 cos
 cos x  1
2
2
Substitute for cos x/2:

2


2


cos
x



 1  cos x 
 2
  cos x
2


 1  cos x  cos x  1
1  cos x
2
Half-angle formulae and applications
EXERCISES
Exercise 1
Use the half-angle formula to evaluate sin 75°.
1  cos150
sin 75  
2

1  0.866
 0.9659
2
Exercise 2
Find the value of sin(α/2) if cos α = 12/13 where
0°< α <90°.

1  cos 
sin

2

1  12
13

2
2
1
26
Exercise 3
2 x
Prove the identity 2 cos
 cos x  1
2
2
 1  cos x 
x
2cos  cos x  2 
  cos x
2
2


2
 2  cos x  cos x  1
Exercise 4
x
Prove the identity 2 sin
 cos x  1
2
2


x
1

cos
x
2
2sin  cos x  2 

2
2


2
2  2cos x
 1  cos x 
 2
 cos x
  cos x 
2
 2 
 1  cos x  cos x  1