Question

How do you use the angle sum or difference identity to find the exact value of `tan(pi/4+pi/3)`?

Answer 1

`tan (pi/4 + pi/3) = -2-sqrt(3)`

Explanation:

From a `1:1:sqrt(2)` right angled triangle (half of a square), we find that `tan(pi/4) = 1`

From a `1:sqrt(3):2` right angled triangle (half of an equilateral triangle), we find that `tan(pi/3) = sqrt(3)`

The sum formulae for `sin` and `cos` are:

`sin(alpha+beta) = sin(alpha)cos(beta)+sin(beta)cos(alpha)`

`cos(alpha+beta) = cos(alpha) cos(beta) - sin(alpha)sin(beta)`

From these we can deduce the sum formula for `tan` (in case you did not know it already):

`tan(alpha+beta) = sin(alpha+beta)/cos(alpha+beta)`

`color(white)(tan(alpha+beta)) = (sin alpha cos beta + sin beta cos alpha)/(cos alpha cos beta - sin alpha sin beta)`

`color(white)(tan(alpha+beta)) = ((sin alpha cos beta + sin beta cos alpha) -: (cos alpha cos beta))/((cos alpha cos beta - sin alpha sin beta) -: (cos alpha cos beta))`

`color(white)(tan(alpha+beta)) = (tan alpha + tan beta)/(1 - tan alpha tan beta)`

Hence we have:

`tan (pi/4 + pi/3) = (tan (pi/4) + tan (pi/3)) / (1 - tan (pi/4) tan (pi/3))`

`color(white)(tan (pi/4 + pi/3)) = (1 + sqrt(3)) / (1 - sqrt(3))`

`color(white)(tan (pi/4 + pi/3)) = ((1 + sqrt(3))(1+sqrt(3))) / ((1 - sqrt(3))(1+sqrt(3)))`

`color(white)(tan (pi/4 + pi/3)) = (1 + 2sqrt(3) + 3) / (1 - 3)`

`color(white)(tan (pi/4 + pi/3)) = (4 + 2sqrt(3)) / (-2)`

`color(white)(tan (pi/4 + pi/3)) = -2-sqrt(3)`