Algebra & Trig, Sullivan & Sullivan
Fourth Edition
Notes:§7.5
Page 1 / 2
§7.5 – Double and Half Angle Formulas
Double angle identities:
sin 2   2 sin  cos
cos2   cos 2   sin 2 
(equivalently, these are also true: (since Cos2θ + Sin2θ = 1) )
cos2   1  2 sin 2 
cos2   2 cos 2   1
2 tan 
tan 2  
1  tan 2 
Practice using these
Find the exact value of sin(2θ), and cos(2θ), each given expression, assuming that 0 ≤θ≤2π
sin    
7
25
,


2
Outline of solution:
1. Construct the triangle (using the Pythagorean Theorem) & the quadrant info
2. Expanding the definitions of sin(2θ), and cos(2θ) so that we've only got sin(θ), and cos(θ)
3. Using the definitions of sin/cos to fill in the various values
4. Plug-and-chug
How do we get these?
Start with sin      cos  sin   sin  cos  , and say that α = β = θ, simplify
sin 2   2 sin  cos
Start with cos     cos  cos   sin  sin  , and say that α = β = θ, simplify
cos2   cos 2   sin 2 
From there, remember that Cos2θ + Sin2θ = 1, we can rearrange to solve for cos, or sin
cos2   1  2 sin 2 
cos2   2 cos 2   1
Algebra & Trig, Sullivan & Sullivan
Fourth Edition
Notes:§7.5
Page 2 / 2
Half angle identities:
1  cos 
 
sin   
2
2
1  cos 
 
cos  
2
2
1  cos 
 
tan   
1  cos 
2
How do we get these?
In both cases, we start with a cos(2θ) formula:
 
For sin  
2
cos2   1  2 sin 2 
 
 
cos 2    1  2 sin 2  
2

2
 
cos   1  2 sin 2  
2
 
cos   1  2 sin 2  
2
 cos   1
 
 sin 2  
2
2
 
For cos 
2
cos2   2 cos 2   1


Define
2
 
 
cos 2    2 cos 2    1
2

2
Simplify (Cancel)
 
cos   2 cos 2    1
2
Move the (-)1 over to the other side
 
1  cos   2 cos 2  
2
Move the (-)2 over to the other side
1  cos 
 
 cos 2  
2
2
Note that these are useful in and of themselves: we'll use them to get a ½ angle formula for tan
Square root of both side
(remember that sin2x is actually (sinx)2)
(remember also that since we started with an identity that we know to already be true, we can
treat it like an equation)
1  cos 
 
 sin 2  
2
2
1  cos 
 
 sin  
2
2
For tan, we start slightly differently:
1  cos 
 
 cos 2  
2
2
Simplify, and voila!
1  cos 
 
 cos 
2
2
1  cos 

sin  
   1  cos 
 
2
  ) tan 2   
tan 2   =
 (Define
 tan 2   

2
2
cos  
 2  1  cos 
 2  1  cos 
2
1  cos 
1  cos 
 
 
 tan   
tan 2   
1  cos 
1  cos 
2
2
2