Notice that
11/12 = 1/6 + 3/4
so that
tan(11π/12) = tan(π/6 + 3π/4)
Then recalling that
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
⇒ tan(x + y) = (tan(x) + tan(y))/(1 - tan(x) tan(y))
it follows that
tan(11π/12) = (tan(π/6) + tan(3π/4))/(1 - tan(π/6) tan(3π/4))
tan(11π/12) = (1/√3 - 1)/(1 + 1/√3)
tan(11π/12) = (1 - √3)/(√3 + 1)
tan(11π/12) = - (√3 - 1)²/((√3 + 1) (√3 - 1))
tan(11π/12) = - (4 - 2√3)/2
tan(11π/12) = - (2 - √3) … … … [A]
When is this homework due?
The center points , so the ones with a full line divide by 2
