Known vector a=(cosα, sin α); b=(cosβ，sinβ)； (1). If α-β=π/6, find the value of a●b; ② If a●b=4/5,

α=π/8, and find the value of tan(α+β).

Solution: (1). a●b = cosαcosβ+sinαsinβ= cos(α-β)= cos(π/6)=√3/2；

(2) If a ● b = cos α cos β+sin α sin β = cos (α-β) = 4/5, then α-β = arccos (4/5), α=π/8 and β=π/8-arccos(4/5).

So α+β = π/8+π/8-arccos (4/5) = π/4-arccos (4/5) ........... (1).

Or α-β=-arccos(4/5), α=π/8, β=α+arccos(4/5)=π/8+arccos(4/5),

So α+β = π/8+π/8+arccos (4/5) = π/4+arccos (4/5) ... (2)

Therefore, tan (α+β) = tan [π/4-arccos (4/5)] = [tan (π/4)-tanarccos (4/5)]/[1+tan (π/4) tanarccos (4/5)]

=[ 1-(3/4)]/[ 1+(3/4)]= 1/7

Where tan arccos (4/5) = [sin arccos (4/5)]/[cos arccos (4/5)] = [√ (1-16/25)]/(4/5) = 3/4.

Or tan (α+β) = tan [π/4+arccos (4/5)] = [tan (π/4)+tanarccos (4/5)]/[1-tan (π/4) tanarccos (4/5)]

=[ 1+(3/4)]/[ 1-(3/4)]=7.

That is, tan(α+β)= 1/7 or 7.