Derive a formula for cos 3θ,

Question

in terms of cos θ and sin θ or just cos θ.?

(4) (a) Derivation: Derive a sum formula for sin(α + β + γ).

 

 

Answer 1

• cos(3θ) = cos(2θ + θ).

Compound angle formula : cos(A + B) = cos A cos B - sin A sin B.

cos(3θ) = cos(2θ + θ)

= cos(2θ) cos(θ) - sin(2θ)sin(θ)

Double angle formula : cos(2A) = 2cos²  A - 1,

                                         sin(2A) = 2 sin A cos A.

= [ (2cos²  θ - 1) cos(θ) ] - [ (2 sin θ cos θ) sin(θ) ]

= 2cos³  θ - cos(θ) - 2sin²  θ cos(θ)

Pythagorean identity : sin²  A + cos²  A = 1.

= 2cos³  θ - cos(θ) - 2(1 - cos²  θ)cos(θ)

= 2cos³ (θ) - cos(θ) - 2 cos(θ) + 2cos³  θ

= 4cos³ (θ) - 3cos(θ).

∴ cos(3θ) = 4cos³ (θ) - 3cos(θ).

• (4).(a).

sin(α + β + γ ) = sin((α + β) + γ).

Compound angle formula : sin(A + B) = sin A cos B + cos A sin B,

                                                cos(A + B) = cos A cos B - sin A sin B.

= [sin(α + β) cos γ] + [cos(α + β) sin γ]

= [(sin α cos β + cos α sin β) cos γ] + [cos α cos β - sin α sin β) sin γ]

= sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ - sin α sin β sin γ.

∴ sin(α + β + γ ) = sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ - sin α sin β sin γ.