Question

What is sin [2tan^-1(3/2)]?

Answer 1

Before proceeding further:
\[\huge \color{green}{\textbf{Welcome To Openstudy...}}\]

Answer 2

thanks

Answer 3

So do you know formula for:
\[2\tan^{-1}(x)\quad ?\]

Answer 4

no I don't.

Answer 5

So here it goes:
\[2\tan^{-1}(x) = \sin^{-1}(\frac{2x}{1 + x^2})\]

Answer 6

So here you have to just convert tan into sine inverse because we have one sine outside the brackets and that will give the answer..

Answer 7

do I have to use that equation to find 2tan^-1(x)?

Answer 8

is there any other way?

Answer 9

See:
you are given with 2tan^-1(x) here in your question x is 3/2.

So firstly we have to convert this tan^-1 into sin^-1 to get our answer

Answer 10

is there a way of doing it using triangles?

Answer 11

See I do some part to explain it more to you:
\[2 \tan^{-1}(x) = \sin^{-1}(\frac{2x}{1 + x^2})\]

Here x = 3/4 so:
\[\sin^{-1}(\frac{2 \times \frac{3}{2}}{1 + (\frac{3}{2})^2})\]

Just solve this.

Answer 12

May be, but I am not aware about that.., Sorry..

Answer 13

And x = 3/2 there..

Answer 14

Might I suggest a way that only involves a few identities that should be relatively simple?

Answer 15

Yep, carry on.

Answer 16

Thank you very much though @waterineyes, i'm just not familiar those formulas yet, and it just isn't the method I was looking for

Answer 17

Well, guys, I should like you to recall that
\[\huge \sin(2x)=2\sin(x)\cos(x)\]

Answer 18

And now, let \(\large x =\tan^{-1}\left(\frac32\right)\)

Answer 19

It is okay @seokhwanahn...

Answer 20

Catch me so far, @seokhwanahn ?

Answer 21

what about the 2 in front?

Answer 22

nevermind, please continue

Answer 23

Yeah, we fixed that...\[\huge \sin(2x)=2\sin(x)\cos(x)\]
\[\large\sin\left[2\tan^{-1}\left(\frac32\right)\right]=2\sin\left[\tan^{-1}\left(\frac32\right)\right]\cos\left[\tan^{-1}\left(\frac32\right)\right]\]

Answer 24

And now, we can do this, as you preferred, with triangles...

Answer 25

So, the tangent of this angle is 3/2, right?