Finding secant with calculator

1. Feb 2, 2012

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

2. Feb 2, 2012

eumyang

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).

3. Feb 2, 2012

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

4. Feb 2, 2012

phyzguy

Since √2 * √2 = 2, then 2/√2 = √2 .

5. Feb 2, 2012

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

6. Feb 2, 2012

SammyS

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

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

7. Feb 2, 2012

bobsmith76

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

8. Feb 2, 2012

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.

9. Feb 2, 2012

SammyS

Staff Emeritus
Actually, neither one gives 2√(2) .

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

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

⟵ This is correct !

10. Feb 2, 2012

bobsmith76

o, don't I feel stupid. thanks

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