DOUBLEANGLE AND
HALF-ANGLE
IDENTITIES
If we want to know a formula for sin 2 x we could use
the sum formula.
sin 2 x  sin  x  x   sin x co s x  co s x sin x
we can trade these places
sin x co s x  sin x co s x  2 sin x co s x
This is called the double angle formula for sine since it
tells you the sine of double x
sin 2 x  2 sin x cos x
Let's try the same thing for co s 2 x
co s 2 x  co s  x  x   co s x co s x  sin x sin x
 cos x  sin x  1  sin x  sin x
2
2
2
 1  2 sin x
2
sin x  1  cos x
2
2
2
cos x  1  sin x
2
2
 cos x  1  cos x   2 cos x  1
2
2
2
This is the double angle formula for cosine but by
substituting some identities we can express it in a
couple other ways.
cos 2 x  cos x  sin x
2
2
Double-angle Formula for Tangent
tan 2 x  tan  x  x  
tan 2 x 
tan x  tan x
1  tan x tan x
2 tan x
1  tan x
2
Summary of
Double-Angle Formulas
sin 2 x  2 sin x cos x
cos 2 x  cos x  sin x
2
2
cos 2 x  1  2 sin x
2
cos 2 x  2 cos x  1
2
tan 2 x 
2 tan x
1  tan x
2
sin x 
4
5
,

Let's draw a
picture.
 x
2
F in d sin  2 x   
24
25
sin 2 x  2 sin x cos x
 4  3 
sin 2 x  2     
 5  5 
4
5
x’
-3
Use triangle to
find values.
x
We can also derive formulas for an angle divided by 2
(called the half angle formula). We’ll do this by using the
double angle formula for cosine that we found.
cos 2  1  2 sin 
2
1  cos 2
 sin 
2
Let’s solve this for sin 

 sin 
2
2
In this formula it is NOT both
+ and - but you must figure
out where the terminal side
of the angle is and put on the
appropriate sign for that
quadrant.
1  cos 2
Now let  = x/2

1  co s x
2
 sin
x
2
We can also derive a half angle formula for cosine in a
similar manner. We’ll do this by using a different version
of the double angle formula for cosine.
cos 2  2 cos   1
2
1  cos 2
 cos 
2
Let’s solve this for cos 

 co s 
2
2
In this formula it is NOT both
+ and - but you must figure
out where the terminal side
of the angle is and put on the
appropriate sign for that
quadrant.
1  co s 2
Now let  = x/2

1  cos x
2
 cos
x
2
Now to derive a half angle formula for tangent, let’s use
the fact that we know that tangent is sine over cosine and
use their half angle formulas.
tan
x
2
sin

co s
x
2

x
tan
2
2

2
x

1  cos 2
1  cos 2
2

1  cos x
1  cos x
Summary
Half-Angle Formulas
sin
cos
tan
x

1  cos x
2
2
x
1  cos x

2
2
x
1  cos x
2

As stated it is NOT both +
and - but you must figure
out where the terminal side
of the angle is and put on
the appropriate sign for
that quadrant.
1  cos x
where the  or - is determined by what quadrant
x
2
is in.
You can also derive identities for
x
the half-angle formulas for tan  
2
in a couple other forms.
sin x
 x  1  cos x
tan   

sin x
1  cos x
2
We could find sin 15° using the half angle formula.
sin

30°
1  cos 
30°
2
2

Since 15° is half of 30° we could
use this formula if x = 30°
15° is in first quadrant
and sine is positive there
so we want the +
3
1
2
sin 15 
2
2
sin 15 
2
2
1
3

2
4
3

2
2
3
sin x 
4

,
5
Find sin
x
2
 x
2

If  is in quadrant II
then half  would be in
quadrant I where sine is
positive
2 5
5
4
sin
x
2
sin
x
2


1  cos x
2
 3
1   
 5
2
Use triangle to
find cosine
value.
1
2
3
8
5 
5 
2
1
5
x’
x
-3
4
5

2
5

2
5
5