Worksheet 4.8 Properties of Trigonometric Functions

We also have
sin 2θ = sin(θ + θ)
= sin θ cos θ + cos θ sin θ
= 2 sin θ cos θ.
We have now established identities 1○ and 2○.
Half Angle Identities
1○: cos 2
2○: sin 2
θ
=
2
1 + cos θ
2
θ
=
2
1 cos θ
2
To prove the half angle identities we begin by rearranging the double angle identities.
Proof of 1○: We take the double angle identity cos 2θ = 2 cos 2 θ − 1 to obtain
2 cos 2 θ = cos 2θ + 1
i.e. cos 2 θ
i.e. cos
=
cos 2θ + 1
2
2
θ
2
= cos θ + 1
2
Proof of 2○: We take the double angle identity cos 2θ = 1 − 2 sin 2 θ to obtain
2 sin 2 θ = 1 − cos 2θ
i.e. sin 2 θ
i.e. sin
=
1 − cos 2θ
2
2
θ
2
= 1 − cos θ
2
We have now established identities 1○ and 2○.
Example 1 : Simplify 2 cos x cos 2x sin 3x − 2 sin x sin 2x sin 3x.
2 cos x cos 2x sin 3x − 2 sin x sin 2x sin 3x
= 2 sin 3x (cos x cos 2x − sin x sin 2x) (factorise)
= 2 sin 3x (cos(x + 2x)) (difference identity)
= 2 sin 3x cos 3x
= sin 2(3x) (double angle)
= sin 6x
9