How do you use the sum or difference identities to find the exact value of tan((23pi)/12)tan(23π12)?

1 Answer
Nov 30, 2016

tan((23pi)/12) = frac (1 - sqrt(3)) (1+ sqrt(3)) tan(23π12)=131+3

Explanation:

First note that (23pi)/12 =2pi-pi/1223π12=2ππ12

and as tan (x+2pi) = tan(x)tan(x+2π)=tan(x), then:

tan((23pi)/12) = tan(-pi/12)tan(23π12)=tan(π12)

Now: 1/12=1/3-1/4112=1314,

So:

tan(-pi/12) = tan(pi/4-pi/3)tan(π12)=tan(π4π3)

Using:

tan(alpha+beta) = frac (tan alpha - tan beta) (1+tan alpha tan beta)tan(α+β)=tanαtanβ1+tanαtanβ

tan((23pi)/12) = frac (tan (pi/4) - tan (pi/3)) (1+ tan (pi/4) tan (pi/3)) tan(23π12)=tan(π4)tan(π3)1+tan(π4)tan(π3)

tan((23pi)/12) = frac (1 - sqrt(3)) (1+ sqrt(3)) tan(23π12)=131+3