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Double-Angle and
Half-Angle Formulas
Double-Angle
Identities
sin 2θ
θ = 2 sinθ
θ cosθ
θ
cos 2θ
θ = cos2θ Š sin2θ = 1 Š 2 sin2θ = 2 cos2θ Š 1
2tanθ
tan 2θ
θ =
1 − tan 2θ
Three Forms of the DoubleAngle Formula for cos2θ
cos 2θ = cos 2 θ − sin 2 θ
cos 2θ = 2 cos 2 θ − 1
cos 2θ = 1 − 2 sin 2 θ
1
Power-Reducing Formulas
1 − cos 2θ
2
1 + cos 2θ
cos 2 θ =
2
1 − cos 2θ
tan 2 θ =
1 + cos 2θ
sin 2 θ =
Example
• Write an equivalent expression for sin4x that does
not contain powers of trigonometric functions
greater than 1.
Solution
 1 − cos 2 x  1 − cos 2 x 
sin 4 x = sin 2 x sin 2 x = 


2
2



1 + cos 2 x 


 1 − 2 cos 2 x + cos 2 2 x   1 − 2 cos 2 x +
2

 = 

4
4

 




 2 − 4 cos 2 x + 1 + cos 2 x  (3 − 3 cos 2 x )
=
=
8
8


Half-Angle Identities
x
sin 2 = ±
1 – cos x
2
x
cos 2 = ±
1 + cos x
2
x
tan 2 = ±
1 – cos x
sin x
1 – cos x
1 + cos x = 1 + cos x = sin x
x
where the sign is determined by the quadrant in which 2 lies.
2
Text Example
Find the exact value of cos 112.5°.
Solution
Because 112.5° = 225°/2, we use the half−angle formula for cos α/2
with α = 225°. What sign should we use when we apply the formula? Because
112.5° lies in quadrant II, where only the sine and cosecant are positive, cos
112.5° < 0. Thus, we use the − sign in the half−angle formula.
cos112.5o = cos
225
2
1 + cos225o
=−
=
2
=−
o
− 2

1+ 
 2 
2
2− 2
2− 2
=−
4
2
Half-Angle Formulas for:
1 − cos α
2
sin α
sin α
α
tan =
2 1 + cos α
tan
α
=
Example
• Verify the following identity:
(sin θ − cosθ ) 2 = 1 − sin 2θ
Solution
(sin θ − cosθ ) 2
= sin 2 θ − 2 sin θ cosθ + cos 2 θ
1 − cos 2θ 1 + cos 2θ
+
− 2 sin θ cosθ
2
2
2
= − 2 sin θ cosθ = 1 − sin 2θ
2
=
3
Double-Angle and
Half-Angle Formulas
4
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