 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 θ2 :
 cos θ = 2 cos2 θ2 − 1.
 On solving this algebraically for cos θ2 , we will have the half-angle

formula for the cosine.

So, on transposing 1 and exchanging sides, we have

 2 cos2 θ2 = 1 + cos θ cos2 θ2 = ½(1 + cos θ) cos θ2 = .

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 θ2 comes from putting 2 = θ in line (3).  On

transposing, line (3) becomes

 2 sin2 θ2 = 1 − cos θ, so that sin θ2 = .

This is the half−angle formula for the sine. Trigonometric identities