Question

Show that `tan(1/2 sin^-1  3/4) = (4 - sqrt(7))/3` and justify why the other value `(4 + sqrt(7))/3` is ignored?

Answer 1

We have `tan(1/2 sin^-1  3/4)`

Let `1/2 sin^-1  3/4` = θ

⇒ `sin^-1  3/4` = 2θ

⇒ sin 2θ = `3/4`

⇒ `(2 tan theta)/(1 + tan^2theta) = 3/4`

⇒ `3  tan theta^2 - 8` and θ ++ 3 = 0

⇒ tan θ = `(8 +- sqrt(64 - 36))/6`

⇒ tan θ = `(8 +- sqrt(28))/6 = (8 +- sqrt(7))/6 = (4 + sqrt(7))/3`

Now `- pi/2 ≤ sin^-1  3/4 ≤ pi/2`

⇒ `(-pi)/2 ≤ 1/2 sin^-1  3/4 ≤  pi/2`

∴ `tan((-pi)/2) ≤ tan(1/2(sin^-1  3/4)) ≤ tan  pi/4`

⇒ `-1 ≤ tan (1/2 sin^-1  3/4) ≤ 1`

⇒ tan θ = `(4 - sqrt(7))/3`  ....`(tan theta = (4 + sqrt(7))/3 > 1, "which is not possible")`