Question

tan22.5*tan37.5*tan82.5

Answer 1

based on a rather little used identity:
tan(A/2) = (1 - cosA)/sinA
and noting that 22.5= 45/2, 37.5 = 72/2, and 82.5 = 165/2

I will skip writing the degree symbol, let it be understood that these are in degrees
we have:
tan 22.5 = (1 - cos45)/sin 45 <----- I know those:
= 1 - √2/2)/√2/2 = (2 - √2)/√2 = √2 - 1

In the same way:
tan37.5 = tan(75/2) = (1 - cos75)/sin75
= <b>(1 - sin15)cos15</b> using complementary trig ratios

tan82.5 = tan(165/2) = (1 - cos165)sin165
= <b>(1 + cos15)/sin15</b> , since cos 165 = -cos15, and sin165 = sin15

tan22.5*tan37.5*tan82.5
= (√2 - 1)*(1 - sin15)cos15) * (1 + cos15)/sin15
= (√2 - 1)( 1 - sin15)(1 + cos15)/(sin15cos15) , but sin15cos15 = (1/2)sin30 = (1/2)(1/2) = 1/4
= 4(√2 - 1)(1 + cos15 - sin15 - sin15cos15)
= 4(√2 - 1)(1 + cos15 - sin15 - 1/4)
= 4(√2 - 1)(3/4 + (cos15-sin15) )

grrrhhhh, but cos 15 - sin 15 = ???

let cos15-sin15 = x
square both sides:
x^2 = cos^2 15 - 2sin15cos15 + sin^2 15 = 1 - sin30 = 1/2
x = √1 / √2 = √2/2
so = 4(√2 - 1)(3/4 + (cos15-sin15) )
= 4(√2 - 1)(3/4 + √2/2) = 4(√2- 1)(3 + 2√2)/4
= ( √2 - 1)(3 + 2√2)

Yeahhh, I hope somebody can see an easier way????