finding secant with calculator

[bobsmith76]

1. The problem statement, all variables and given/known data

what is sec(pi/4)
what is sec(0)



3. The attempt at a solution

First let me make sure that cos^-1, acos and sec, they are all the same right?

I put in cos^-1(pi/4) in my calculator and the answer I get is .667, which I think is the same as sec (pi/4). The book says the answer is √2

[eumyang]

1. The problem statement, all variables and given/known data

what is sec(pi/4)
what is sec(0)



3. The attempt at a solution

First let me make sure that cos^-1, acos and sec, they are all the same right?
No! They are not the same. The first two, cos-1 and acos (do you mean arccosine?) are the inverse cosine functions. Put it simply, if cos(π/6) = √3/2, then cos-1(√3/2) = π/6.

You are confusing these with secant, which is the reciprocal of cosine. In order to evaluate secant on the calculator, you need to type
1/cos(your angle).

[bobsmith76]

But I still don't see why sec(pi/4) is √2, it should be 2/√2 and my book clearly says it's √2

[phyzguy]

Since √2 * √2 = 2, then 2/√2 = √2 .

[bobsmith76]

That doesn't make sense. cos (pi/4) = √2/2. If you take the inverse, which is the secant, then it's 2/√2, not sqaure root of 2

[SammyS]

You could do the following subtraction:
\displaystyle \sqrt{2}-\frac{2}{\sqrt{2}}
The answer is zero.

Or the following multiplication:
\displaystyle \frac{\sqrt{2}}{1}\cdot\frac{\sqrt{2}}{\sqrt{2}}

[bobsmith76]

Sammy, all that gives 2√2, the book clearly says otherwise

http://i87.photobucket.com/albums/k137/kylefoley76/Screenshot2012-02-02at53653PM.png

[phyzguy]

What we're trying to tell you is that 2/√2 and √2 are just different ways of writing the same number. They are equal. They are the same. You and the book are both right.

[SammyS]

Sammy, all that gives 2√2, the book clearly says otherwise

Actually, neither one gives 2√(2) .

\displaystyle \sqrt{2}-\frac{2}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}}{\sqrt{2 }}-\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{2}}=0

\displaystyle \frac{\sqrt{2}}{1}\cdot\frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}\sqrt{2}}{\sqrt{2}}=\frac{2}{\sqrt{2 }}

http://i87.photobucket.com/albums/k137/kylefoley76/Screenshot2012-02-02at53653PM.png ⟵ This is correct !

[bobsmith76]

o, don't I feel stupid. thanks