Question

DIFFERENTIATING inverse trig functions. :)
Question http://prntscr.com/61cfl0. I just need help with proving LOL

Answer 1

I've already done Part A which just gives you zero but i'm not sure how to do Part B :/

Answer 2

@butterflydreamer I don't see the question.

Answer 3

http://prntscr.com/61cfl0

Answer 4

can you see it? :O

Answer 5

Yep but I don't know this lol Sorry:(

Answer 6

oh haha. it's okay :) thanks for stopping by

Answer 7

Np if you have any Geometry questions just tag my name :)

Answer 8

ok

bring either sin^-1 or cos^-1 to other side

Answer 9

you want to prove it using derivatives

Answer 10

oh alright. I just thought that if they ask you to prove.. usually you do that "LHS= ...
" and then in the end it's meant to equal RHS or something? :/
But wouldn't we also be using part A to answer Part B??

Answer 11

ok, not sure how to use derivatives,
but you can use triangle method to prove that identity

Answer 12

make a right triangle
label an acute angle as 'x'

find sin x
and sin (90-x)
and cos x

Answer 13

or you can use the sin(a+b) formula too,
if you want to do it algebraically instead of making triangle...

Answer 14

O_O......
what is a and b in sin (a + b) ?? feeling confused LOL

Answer 15

sin^-1 x = a
cos^-1 x =b

Answer 16

so
sin a = x
cos b = x

Answer 17

hmm..... that way seems so complicated D:
I was thinking.. since Part A we found the gradient =0.. does that mean sin^-1x + cos^-1x is a constant?

Answer 18

yes you can conclude that
it is definitely a constant

but how will u prove that constant is pi/2

Answer 19

well if the function was a constant...
Couldn't we substitute in any value for x ? or something along those lines xD LOL i have no idea actually..