Question

Find sec(2Arcsin (-1/3))

Answer 1

If $\theta=\sin^{-1}\left(-\dfrac13\right)$, then $\sin\theta=-\dfrac13$. We want to find $\sec2\theta$. Recall that

$\sec2\theta=\dfrac1{\cos2\theta}$

and by the double angle identity,

$\cos2\theta=1-2\sin^2\theta$

So we have

$\cos2\theta=1-2\left(-\dfrac13\right)^2=\dfrac79$

$\implies\sec2\theta=\boxed{\dfrac97}$

Answer 2

Think of arcsin (-1/3) as an angle, and call it "theta."  The the original problem becomes "find sec [ 2 theta ]"
                                                       1                                    1
which is the same as     "find ------------------ " or --------------------------------------
                                                cos[2 theta]          (cos theta)^2 - (sin theta)^2

Recall that theta = the angle whose sine is -1/3.  Thus, sin theta = -1/3 and cos theta = 2sqrt(2) / 3.

So we now have: 

                   1
-------------------------------------------
   2 sqrt(2)                     -1
[------------------- ]^ 2 - [ ------- ]^2
           3                           3

which boils down to:

          1                   1
----------------- = ---------------- = 9/7 (answer)
   8         1               7/9
------ - --------
   9          9