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Derive a formula for cos 3θ,

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

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

 

 

asked Jul 10, 2014 in ALGEBRA 1 by anonymous

1 Answer

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  • 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.

answered Jul 10, 2014 by lilly Expert

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