Question

cot(sin^-1(-2/3))
Could someone please show me the steps to solving this without a calculator?

Answer 1

Ahhhh, this is an interesting problem. Well, do you know how to use the fundamental ID, along with knowledge of the quadrant that an angle is in, to find the cos(x), given the sin(x)?

Answer 2

That is, if I said that

sin(x)=# and x is in quadrant 4, find cos(x)

would you know how to do that?

Answer 3

yup :)

Answer 4

ok, now look at the "inner" part of the expression above.

\[\Large \sin^{-1}(-2/3)\] is an angle x such that:

sin(x) = ?
and
sin(x) is in what quadrant?

Answer 5

(you know the answer to both, by the definition of the inverse sine function) :)

Answer 6

So once you answer those 2 questions, you will be able to find cos(x) using the fundamental ID.

That means you will have sin(x) and cos(x), from which I bet you know how to find cot(x). :)

Answer 7

how do you find arcsin (-2/3) without a calculator though?

Answer 8

That is the beauty of this problem. You don't NEED to find it.

Let \(\Large x=\sin^{-1}(-2/3)\)

Remember what that inverse function means... it just means "the angle that has that sine value, and is in the range of arcsin"

Answer 9

So because the range of arcsin is limited, you KNOW the quadrant that x lies in.

And you have the sin(x), right? You don't NEED x, you just need to know sin(x) and cos(x) to get cot(x).

You are given sin(x), and enough info to find cos(x) with Fund'l ID. :)

Answer 10

Because with \(\Large x=sin^{-1}(-2/3)\), the problem is simply find \(\Large cot(x)\).

Answer 11

Wait i think i get it. The answer is \[-\sqrt{5/2}\].

Answer 12

whoops i mean \[\sqrt{5}/-2\]

Answer 13

and also thank you very much :)

Answer 14

Haha... was just about to say that. :) That's what I get too! :)

Answer 15

You're quite welcome, happy to help. I liked this problem, will have to remember it. :)