Why are Cosine and Secant Even Functions?

[CrossFit415]

Homework Statement

Why is Cosin and Secant even? Cos (-t) = cos t, Sec (-t) = sec t
Why don't they equal - sec t instead like the rest of the functions? Thanks

Homework Equations

The Attempt at a Solution

 

[eumyang]

You could look at their graphs. Even functions have graphs that are symmetric at the y-axis -- what appears on the left side of the y-axis appears again as a "mirror" on the right side of the y-axis. The graphs of the rest of the trig functions do not exhibit this behavior.

 

[CrossFit415]

Hmm I somewhat kind of get it now. Thanks a lot!

 

[HallsofIvy]

How, exactly, are you defining "cosine". Probably you are using something like "Given a number t, measure a distance t around the circumference of the unit circle, starting at (1, 0). cos(t) is the x coordinate of the ending point." From that it should be clear that if t> 0 takes you to the point (x, y), -t takes you to (x, -y). x= cos(t)= cos(-t), y= sin(t), -y= sin(-t).

 

[romsofia]

The reason why cosine is an even functions is because if you expand cosine via power series you'll get polynomials that are only even: 1-X²/2!+X⁴/4!-X⁶/6!+X⁸/8!...

 

[eumyang]

The reason why cosine is an even functions is because if you expand cosine via power series you'll get polynomials that are only even: 1-X²/2!+X⁴/4!-X⁶/6!+X⁸/8!...

I don't think this is helpful to the OP, because, unless I'm mistaken, he/she hasn't seen power series yet.

 

[romsofia]

I don't think this is helpful to the OP, because, unless I'm mistaken, he/she hasn't seen power series yet.

I thought about that before posting, but I'm sure it'll be useful to see why it's really an even function.

 

[HallsofIvy]

In fact, it is possible to define cos(x) in terms of its Taylor series. For such a definition, romsofia's response is perfect. But we don't know what kind of response is appropriate until we know exactly how the OP is defining cosine and secant.