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

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

becomes

θ 2

:

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

Copyright © 2021 Lawrence Spector

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