Trigonometry

Proof of the double-angle and half-angle formulas

Double-angle formulas

Proof

The double-angle formulas are proved from the sum formulas by putting β = .  We have

sin 2 = sin ( + ) = sin cos + cos sin
 
  = 2 sin cos .
 
cos 2 = cos ( + ) = cos cos − sin sin
 
cos 2 = cos² − sin².   .  .  .  .  .  . (1)

This is the first of the three versions of cos 2.  To derive the second version, in line (1) use this Pythagorean identity:

sin² = 1 − cos².

Line (1) then becomes

cos 2 = cos² − (1 − cos²)
 
  = cos² − 1 + cos².
 
cos 2 = 2 cos² − 1.  .  .  .  .  .  .  .  .  .  (2)

To derive the third version, in line (1) use this Pythagorean identity:

cos² = 1 − sin².

We have

cos 2 = 1 − sin² − sin²;.
 
cos 2 = 1 − 2 sin².  .  .  .  .  .  .  .  .  .  (3)

These are the three forms of cos 2.

Half−angle formulas

 .  .  .  .  .  .  .  (2')

 .  .  .  .  .  .  .  (3')

Whether we call the variable θ or does not matter.  What matters is the form.  

Proof

Now, is half of 2.  Therefore, in line (2), we will put 2 = θ, so that

   becomes  θ
2
:
cos θ = 2 cos² θ
2
 − 1.
On solving this algebraically for cos  θ
2
, we will have the half-angle

formula for the cosine.

So, on transposing 1 and exchanging sides, we have

2 cos² θ
2
= 1 + cos θ
 
cos² θ
2
= ½(1 + cos θ)
 
cos  θ
2
= .

This is the half-angle formula for the cosine.  The sign ± will depend on the quadrant of the half-angle.  Again, whether we call the argument θ or does not matter.

Notice that this formula is labeled (2') -- "2-prime"; this is to remind us that we derived it from formula (2).

The formula for sin  θ
2
 comes from putting 2 = θ in line (3).  On

transposing, line (3) becomes

  2 sin² θ
2
= 1 − cos θ,
  so that
 
  sin  θ
2
= .

This is the half−angle formula for the sine.


Trigonometric identities


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