Question

Am I right?

Answer 1

\[\sec(-x)=-\sec(x)\]

Answer 2

No, this is like saying:

(-x)^2=-(x^2)

Answer 3

How? I really thought it's an identity.

Answer 4

false

Answer 5

just check a number
\[sec(-3)\ne
-sec(3)\]

Answer 6

http://math2.org/math/trig/identities.htm

Answer 7

Cosine is an even function, therefore secant is an even function. You're probably thinking:

-sin(x)=sin(-x)

Answer 8

http://math2.org/math/trig/identities.htm

Answer 9

My bad!

Answer 10

its even not odd:)

Answer 11

Can you explain then, why is it so that\[\sin(-x)=-\sin(x)\]and\[\sec(-x)=\sec(x)\]

Answer 12

What do you mean even not odd? 4 not 7?

Answer 13

look at the graph
its sort of like (-x)^2 = x^2
and (-x)^3 = -x^3

Answer 14

no
a function is even if f(-x) = f(x)

Answer 15

a function is odd if f(-x) - -f(x)

Answer 16

How is sec odd?

Answer 17

Well think of the graphs. By "even" and "odd" functions we are talking about the symmetry of the function.

Even functions like x^2 or cos(x) are exactly the same when reflected over the y-axis.

Odd functions like x, x^3, or sin(x) are exactly the same when rotated 180 degrees around the origin.

Answer 18

even functions are symmetric across the y axis

Answer 19

its not, its even

Answer 20

cos(x) is even, so 1/cos(x) must be even. And remember, that's what secant is.

Answer 21

yeah. I meant to ask even....
I still don't get why are sin,csc and tan,cot are even and cos,sec are odd

Answer 22

http://www.wolframalpha.com/input/?i=y%3Dcosx%20and%20y%3Dsec%20x&t=crmtb01

A picture helps always. So you can see if you plug in 2 or -2 into either of those you will always get to the same height.

Answer 23

look at the graph

Answer 24

even is symmetric across the y axis, odd is symmetric across the origin.

Answer 25

OK, nvm, I'll just always go with the identities w/o knowing why.
id 've to always ky.

Answer 26

I don't have to....
(explaining what I wrote)