Evaluating inverse trig function

**tom ato** · Apr 23rd 2009, 06:22 AM

dv/dt = 3/[t((t^2)-1)] t>0, v(2) = 0

dv = integral 3/[t((t^2)-1)] dt

v = 3sect^-1 + C

3sec(-2)^-1 = -C

How do I evaluate inverse secant?! Is there any simple rule to go by just looking at the unit circle? My teacher says the answer is c = -pi.

**Soroban** · Apr 23rd 2009, 07:36 AM

Hello, tom ato!

My answer is: . $-{\color{red}2}\pi$

$3\sec^{-1}(\text{-}2) \:=\: -C$

How do I evaluate inverse secant?

We are expected to reason like this . . .

Let $\theta \:=\:\sec^{-1}(\text{-}2)$
. . Then $\theta$ is the angle whose secant is -2.

Hence, $\theta$ is the angle whose cosine is $\text{-}\tfrac{1}{2}$

The "first" such angle is: . $\theta \:=\:\frac{2\pi}{3}$

Therefore: . $\sec^{-1}(\text{-}2) \;=\;\frac{2\pi}{3}$

**tom ato** · Apr 23rd 2009, 09:07 AM

Thanks. And I just realized that the answer should be negative pi because I'm supposed to take the inverse secant of 2 (I had put -2)

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