Question

Would someone mind helping me simplify the expression:

sin( tan^-1 (x) ) or sin(arctan x)

Answer 1

\[\sin (\tan^{-1} x)\]
Is this your equation?

Answer 2

Yep

Answer 3

let tan^-1(x) =u

Answer 4

Ok, and then x = tan (u)?

Answer 5

yes

Answer 6

where u is between -pi/2 and pi/2 right?

Answer 7

right because of tangent?

Answer 8

yes

Answer 9

so then you have sin(u) between -pi/2 and pi/2

Answer 10

right

Answer 11

so then the hypotenuse is gonna be sqrt(x^2+1)

Answer 12

How so? Are you using a right triangle then applying the pythag theorem based off of x/1 = tan(u)?

Answer 13

u got it

Answer 14

is there any other method aside using a right tri or is that just the easiest?

Answer 15

there are other methods but this the one i know sorry

Answer 16

It's alright. Thanks anyway.

Answer 17

np

Answer 18

\[\large\rm \sin\left[\color{orangered}{\arctan(x)}\right]\]Figure it out jtug? :)

Answer 19

Yes! So u = arc tan x so x = tan u and we know that x = x/1 so

x/1 = tan u meaning that x and 1 are opp/adj to tan which we can then use to find
sin(u) BECAUSE u = arc tan x which = our original eq sin [arctan(x)]
so sin(u) = opp/hyp which we dont have hyp yet meaning we use pythag theorem because we can use a right triangle which c^2 = x^2 + 1 so c = sqrt x^2+1 which gives us our hyp so we're left with x/[sqrt(x^2+1)]

:D

Answer 20

Thanks for asking btw.

Answer 21

Calc 2 is beautiful so far. difficult, but beautiful.

Answer 22

ya fun stuff :))