Given tan(α)=−6/sqrt(13) and π/2<α<π, find the exact value of tan(α/2).

Answer 1

Hi Pug,

For this problem I used this identity

tan(a/2) = (-1 ± √(1 + (tan(a))²) / tan(a)

tan(a/2) = (-1 ± √(1 + (-6 / √13)²)) / -6 / √13

tan(a/2) = (-1 ± √(1 + 36 / 13 ) / -6 / √13

tan(a/2) = (-1 ± √(49 / 13 ) / -6 / √13

multiply by √13 / √13

tan(a/2) = (-√(13) ± √(49)) / -6

tan(a/2) = (-√13 ± 7) / -6

get rid of the negative

tan(a/2) = (√13 ± 7) / 6

so the positive answer is

a = 2 * tan^-1((√13 + 7) / 6 )

a = 121°

and

a = 2 * tan^-1((√13 - 7) / 6 )

a = -59°

pi/2 to pi includes 121° but not 59°

so the correct answer is

tan(a/2) = (√13 + 7) / 6

I hope this helps