Question

I don't understand any of the inverse trig functions.. What are they?

Answer 1

What is this right triangle thing

Answer 2

cot(arcsinx)
why is this

Answer 3

If you have an angle, you can find the value of that angle using a trig function.

But what if you're given the value? Then you use the inverse trig function to find the angle.

For example:

we know this \(\sf sin~30^o = \frac{1}{2}\)

But what if we are given this:

\(\sf sin~x^o=\frac{1}{2}\)

We solve it like this:

\(\sf x^o = sin^{-1}\frac{1}{2}\)

And plugging it in to a calc, we get

\(\sf x^o = 30^o\)

sin^(-1) is the same thing as arcsin

and the calc will only give one value, but sin 150 also equals to 1/2 :b

Answer 4

What is this whole right triangle thing?

And there's things like sin(tan^-1x/3)

:O

Answer 5

You might need to elaborate on what you mean by "this whole right triangle thing"?

Answer 6

Can't elaborate when I don't even know what this is

Answer 7

From this diagram, you should immediately know that \[\large \tan \theta = \frac{ x }{ 3 }\]which means that \[\large \theta = \tan^{-1} \frac{ x }{ 3 }\]

Answer 8

What happens next?

Answer 9

I'm assuming you understand all that.

Remember you had \[\large \sin\left( \tan^{-1}\frac{ x }{ 3 } \right)\]
Now look at what I did above.

Answer 10

Specifically look at the last line... hopefully you notice we can do this\[\large \sin\left( \tan^{-1}\frac{ x }{ 3 } \right) = \sin \theta\]

Answer 11

so I have to visualize that tan(x/3) triangle first, then figure out the sin of that?

Answer 12

Exactly!

Answer 13

hmp hmp. I think I got that part!

Answer 14

solve for the missing side labeled ???

Answer 15

Don't try to visualize it, you'll make more mistakes. Draw it out and it'll be easier to see :)

Answer 16

Do I know that's a 60 30 90 triangle?
What type of triangle is that?

Answer 17

when they give you something like cot(arcsinx), it will be a right triangle. We won't know the angles since it will always vary depending on the question but it will be a right triangle. :)

Answer 18

How do I figure out the hypot. in that picture?

Answer 19

\[\Large a^2 + b^2 = c^2\]

Answer 20

\[\sqrt{9+x^2}\]

Answer 21

Right. Now remember you needed to find sin theta
\[\large \sin\left( \tan^{-1}\frac{ x }{ 3 } \right) = \sin \theta = ?\]

Answer 22

I can only solve this by looking at the triangle :(
\[\frac{ x }{ \sqrt{9+x^2} }\]

Answer 23

\[\Large \checkmark \]Why the sadface? \[\large \sin\left( \tan^{-1}\frac{ x }{ 3 } \right) = \sin \theta = \frac{ x }{ \sqrt{9+x^2} }\]

Answer 24

Can't do it in my head.
Can't even visualize the unit circle in my head.

Answer 25

What do you do when... It just tells you to evaluate arcsec2?
What is this..? Why is there so many, when do I draw the triangle and when do I draw the inverse graph etc etc. There's also a "relation" and a "function", where the relation is represented with the capital letter, for instance, Cos^-1.
There's so many.

Answer 26

Why would you need to do any of this in your head?

You have a lot of questions. Pick one and make a new question.

Answer 27

Well, what's the most important one to know?

And I can't draw the unit circle out every time I solve something. That'll take out more than half of my time.

Answer 28

Can we start with... "arcsec2"?
Or something similar..

Answer 29

Yeah but I mean post a whole new question.