Respuesta :

altavistard
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
LammettHash

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

[tex]\sec2\theta=\dfrac1{\cos2\theta}[/tex]

and by the double angle identity,

[tex]\cos2\theta=1-2\sin^2\theta[/tex]

So we have

[tex]\cos2\theta=1-2\left(-\dfrac13\right)^2=\dfrac79[/tex]

[tex]\implies\sec2\theta=\boxed{\dfrac97}[/tex]

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