Question

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

Answer 1

If
`\theta=\sin^(-1)\left(-\frac13\right)`, then
`\sin\theta=-\frac13`. We want to find
`\sec2\theta`. Recall that

`\sec2\theta=\frac1{\cos2\theta}`

and by the double angle identity,

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

So we have

`\cos2\theta=1-2\left(-\frac13\right)^2=\frac79`

`\implies\sec2\theta=\boxed{\frac97}`

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