192tan-1 x = tan-1 {2x / (1-x2)}.....if -1≤x≤1
= tan-1 {2x / (1-x2)} + π.......if x>1
= tan-1 {2x / (1-x2)} -Ï€......if x<-1
also remember that sin2θ = 2 tan θ/(1+tan2 θ)
6 Answers
19
39Going by those formulae, put x = tanθ such that θ = tan-1x.
So y = 2θ + sin-1(2tanθ1+tan2θ)
=> y = 2θ + sin-1sin2θ
=> y = 2θ + 2θ
=> y = 4θ
=> y = 4tan-1x
Now 0 < y4 < pi2, as is the principal domain for tan-1x.
Thus 0 < y < 2pi is the answer.
341Shouldnt it be \left(-\frac{3 \pi}{4}, \frac{3 \pi}{4} \right)?
If you put x = \tan \theta, where -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}then y is given by
\begin{cases} & -(\pi + \theta) \text{ if } -\frac{\pi}{2} \le \theta \le -\frac{\pi}{4} \\ & 3 \theta \text{ if } -\frac{\pi}{4} \le \theta \le \frac{\pi}{2} \\ & \pi - \theta \text{ if } \frac{\pi}{4} \le \theta \le \frac{\pi}{2} \end{cases}