Question

Isn't arcsec=cos? Since sec=1/cos the inverse of sec would be 1/(1/cos) which would be cos??

Answer 1

@Clyde101 arcsecant is not the same as cosine. I think you are confusing the "-1" exponent of arcsecant with reciprocals when the "-1" in fact indicates the inverse function and not the reciprocal. To clear things up, observe this:\[\bf \sec(x) =\frac{1}{\cos(x)} \ \ and \ \ [\sec(x)]^{-1} =\frac{1}{\sec(x)}=\cos(x)\]\[\bf arcsec(x)=\sec^{-1}(x) \ne [\sec(x)]^{-1}=\cos(x)\]

Answer 2

The -1 exponent outside the brackets indicates cos(x) and it also indicates that it's the reciprocal of sec(x). However the -1 exponent inside indicates the inverse function of sec(x) which is not the same as the reciprocal cos(x).

Answer 3

OHHH ok, so then I need help with a problem...how would I solve y=arcsec(-1)

Answer 4

So since arcsec is the inverse function of secant, we can re-write the question like this:\[\bf y=\sec^{-1}(-1) \implies \sec(y)=\frac{ 1 }{ \cos(y) }=-1\]Can you solve for it now?

Answer 5

pi?

Answer 6

For what value of y will 1/cos(y) result in -1. Notice that by re-arranging we get:\[\bf \cos(y)=-1\]Which 'y' would give us -1 as the answer?

Answer 7

That is correct.

Answer 8

And that's it. You got the answer.

Answer 9

but cos is pi for this value as well...so why doesn't arcsec =cos always?

Answer 10

No nonono they are not the same thing.

Answer 11

on the unit circle...pi=cos-1

Answer 12

Here, look at the graphs of both functions here: https://www.desmos.com/calculator/rkz053safx

Answer 13

Do they look the same?

Answer 14

not at all no...so why is the value the same on the unit circle in this instance?

Answer 15

They are not. They have just 1 point of intersection at x = 1.108 and that's the only place they are ever the same.

Answer 16

oh ok i get that

Answer 17

Good. One thing to remember though, is that with inverse functions, the domain/range is usually limited in the cases of arccos(x), arcsin(x) and arcsec(x) and a few others. So when you do something like:\[\bf y=\cos^{-1}(x) \implies \cos(y)=x\]And you are given a value for x and you must get the value for y, you shouldn't forget that when you solve for y using cos(y) = x, that the value of y u find must be in the range of the inverse function. Cos(x) has a range of 1 to -1 but arccos(x) is the inverse, so the domain and range switch and to make sure that it remains a function (passes vertical line test), the domain gets restricted to -1 to 1 and range becomes 0 to pi.

Answer 18

So when you solve for y for a certain inverse trig function, you must keep it's domain/range in mind.

@Clyde101

Answer 19

ok I will thanks that really helps alot! online pre-calc honors isn't easy on your own :)

Answer 20

haha yep i will