Question

Question #48a63

Answer 1

• (1 + sqrt2)

Explanation:

Call tan (5pi)/8 = tan t
Trig table and unit circle give:
`tan 2t = tan ((10pi)/8) = tan ((5pi)/4) = tan (pi/4 + pi) =`
`tan (pi/4) = 1`
Use trig identity: `tan 2t = (2tan t)/(1 - tan^2 t)`
In this case, we have:
`(2tan t)/(1 - tan^2 t) = 1`
`2tan t = 1 - tan^2 t`
`tan^2 t + 2tan t - 1 = 0`
Solve this quadratic equation for tan t, using the improved quadratic formula (Socratic Search):
`D = d^2 = b^2 - 4ac = 4 + 4 = 8` --> `d = +- 2sqrt2`
There are 2 real roots:
`tan t = -b/(2a) +- d/(2a) = - 2/2 +- (2sqrt2)/2 = - 1 +- sqrt2`
Since `tan ((5pi)/8)` is negative (Quadrant II), there for;
`tan t = tan ((5pi)/8) = - (1 + sqrt2)`
Check by calculator:
`tan ((5pi)/8) = tan (112.5) = - 2.414`
`- ( 1 + sqrt2) = - 2.414` . OK