f.4 compound angles

given that sin3θ = 3sinθ - 4sin³θ and cos3θ = 4 cos³θ - 3cosθ, prove the following identities.

(i) sin3θsin³θ+cos3θcos³θ = cos³2θ
LHS
= sin3θsin³θ+cos3θcos³θ
= (3sinθ - 4sin³θ)sin³θ + (4 cos³θ - 3cosθ)cos³θ
= 3sin4θ - 4sin6θ + 4 cos6θ - 3 cos4θ
= 3 sin4θ - 3 cos4θ + 4 cos6θ- 4sin6θ
RHS
= cos³2θ .................... ..................us e the formula: cos2θ = cos²θ - sin²θ
=(cos²θ - sin²θ)³
= cos6θ - 3cos4θsin²θ + 3cos²θ sin4θ - sin6θ
= cos6θ - 3cos4θ(1-cos²θ) + 3(1-sin²θ)sin4θ - sin6θ
= cos6θ - 3cos4θ + 3cos6θ) + 3sin4θ - 3sin6θ - sin6θ
= 3 sin4θ - 3 cos4θ + 4 cos6θ- 4sin6θ
LHS = RHS, ∴ sin3θsin³θ+cos3θcos³θ = cos³2θ


(ii) (cos3θ+2cos2θ+7cosθ+6)/(8cosθ-2sinθsin2θ)=1+secθ
LHS
= (cos3θ+2cos2θ+7cosθ+6)/(8cosθ-2sinθsin2θ)
= [(4 cos³θ - 3cosθ) +2 (cos²θ - sin²θ) + 7cosθ+6] / [8cosθ-2sinθ(2sinθcosθ)]
= {4 cos³θ + 4cosθ +2 [cos²θ - (1- cos²θ)] +6} / (8cosθ-4cosθsin²θ)
= [4 cos³θ + 4cosθ +2 (2cos²θ - 1) +6] / [8cosθ-4cosθ(1-cos²θ)]
= (4 cos³θ + 4cosθ +4 cos²θ - 2 +6) / (8cosθ- 4cosθ + 4 cos³θ)
= (4 cos³θ + 4cosθ + 4cos²θ + 4) / (4cosθ + 4cos³θ)
= [(4cosθ (cos²θ+ 1) +4 (cos²θ + 1)] /[4cosθ (1+ cos²θ)]
= [4 (1+ cos²θ ) (cosθ+ 1) ] /[4cosθ (1+ cos²θ)]
= (cosθ + 1)/cosθ
= 1 + (1/cosθ)
= 1 + secθ
= RHS

∴ (cos3θ+2cos2θ+7cosθ+6)/(8cosθ-2sinθsin2θ)=1+secθ

P.S.: use the formula:　cos2θ = cos²θ - sin²θ　&　sin2θ = 2sinθcosθ