Trigonometry Double Angles

DOUBLE-ANGLE IDENTITIES

These are 4 more identities, known as the double angle identities, that can be derived from
the compound Identities:
(They are on the formula sheet).

𝐬𝐢𝐧 𝟐 𝐀 = 𝟐 𝐬𝐢𝐧 𝐀 𝐜𝐨𝐬 𝐀
𝐜𝐨𝐬 𝟐 𝐀 = 𝐜𝐨𝐬 𝟐 𝐀 − 𝐬𝐢𝐧𝟐 𝐀
𝐜𝐨𝐬 𝟐 𝐀 = 𝟐 𝐜𝐨𝐬𝟐 𝐀 − 𝟏 These 3 are all variations of the same identity

𝐜𝐨𝐬 𝟐 𝐀 = 𝟏 − 𝟐 𝐬𝐢𝐧𝟐 𝐀
Here is where they come from:
(i) cos 2A = cos (A + A)
= cos A . cos A – sin A . sin A compound angle identity
= cos2 A – sin2 A (one form)
= cos2 A – (1 – cos2 A) square identity
= 2 cos2 A – 1 (another form)
= 2(1 – sin2 A) – 1 square identity
= 1 – 2 sin2 A (yet another form)

(ii) sin 2A = sin (A + A)
= sin A . cos A + cos A . sin A compound angle identity
= 2 sin A . cos A

Using Double-Angle Identities:

e.g. 1 Simplify (sin A − cos A)2
Solution:
(sin A − cos A)2
= sin2 A − 2 sin A cos A + cos2 A
= 1 − sin 2A square identity; double angle identity

2. Prove that sin 3θ = 3 sin θ − 4 sin3 θ
Solution:
LHS = sin 3θ
= sin(2θ + θ)
= sin 2θ . cos θ + cos 2θ . sin θ compound angle identity
= 2 sin θ cosθ cos θ + (1 − 2 sin2 θ) sin θ double angle identities
= 2 sin θ cos 2 θ + sin θ − 2 sin3 θ
= 2 sin θ (1 − sin2 θ) + sin θ − 2 sin3 θ square identity
= 2 sin θ − 2 sin3 θ + sin θ − 2 sin3 θ
= 3 sin θ − 4 sin3 θ

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