tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y))
substitute x=y to get
tan(2x) = 2tan(x)/(1-tan²(x))
Substitute tan(2x)=1 and solve for tan(x).
Reject the negative root knowing that tan(x)≥0 for 0<x<π/2.
Given that tan 45=1, use tan(x+y) to show that tan 22.5= /2 - 1.
(/is square root sign
2 answers
tan(x+y) = (tanx + tany)/(1 - tanxtany)
so tan 45
= tan(22.5 + tan22.5)/(1 - tan22.5tan22.5)
= 2tan22.5)/1 - tan^2 22.5)
let tan22.5 = y for ease of typing
1 = 2y/(1 - y^2)
1 - y^2 = 2y
y^2 + 2y - 1 = 0
y^2 + 2y = 1 , by completing the square
y^2 + 2y + 1 = 1+1
(y+1)^2 = 2
y+1 = √2
y = √2-1
tan22.5° = √2 - 1
so tan 45
= tan(22.5 + tan22.5)/(1 - tan22.5tan22.5)
= 2tan22.5)/1 - tan^2 22.5)
let tan22.5 = y for ease of typing
1 = 2y/(1 - y^2)
1 - y^2 = 2y
y^2 + 2y - 1 = 0
y^2 + 2y = 1 , by completing the square
y^2 + 2y + 1 = 1+1
(y+1)^2 = 2
y+1 = √2
y = √2-1
tan22.5° = √2 - 1
2 answers