We have,
sin4α+4cos4β+2=42sinαcosβ
Applying AM≥GM, we get
(4sin4α+4cos4β+1+1)≥(4sin4α⋅cos4β⋅1⋅1)41
⇒sin4α+4cos4β+2≥8sinαcosβ
And, AM=GM⇒numbers are equal and positive
⇒sin4α=4cos4β=1
∴sin4α=1⇒α=2π and
cosβ=21⇒β=4π
Hence,
cos(α+β)−cos(α−β)
=−2sinαsinβ
=−2×1×21=−2