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.

Printable View

- Apr 23rd 2009, 06:22 AMtom atoEvaluating inverse trig function
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. - Apr 23rd 2009, 07:36 AMSoroban
Hello, tom ato!

My answer is: .

Quote:

How do I evaluate inverse secant?

We are expected to reason like this . . .

Let

. . Then is the angle whose secant is -2.

Hence, is the angle whose*cosine*is

The "first" such angle is: .

Therefore: .

- Apr 23rd 2009, 09:07 AMtom ato
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)