Question

Part 1: If y=arcsin x, find cos y and tan y.

Part 2: If y=arctan x, find cos y and sin y.

**How do you start this problem off? Thanks a bunch!! :)

Answer 1

Draw a right triangle using this, make one angle y

y=arcsin x,
then
siny = x or siny = x/1

Answer 2

Then find cosy and tany from the triangle.

Same thing for part 2

Answer 3

okay, so i draw the triangle like this right? (attached)

what do i do next? :/ not sure how to do the arcsin x of it :(

Answer 4

oh whoops it attached sideways haha :P

Answer 5

y=arcsin x,
then
siny = x or siny = x/1

Answer 6

okay.. did i do it right? :/

Answer 7

Well that side should just be x. Now find the remaining side.

Answer 8

okay so should it be

x and then 1 and would the remaining side be (sq.rt. 1- x^2) ?

Answer 9

Yep now find cosy and tany.

Answer 10

how do i do that? :/

Answer 11

You know how to do it if you've gotten this far ;)

Answer 12

okay :P lemme try to remember hehe one sec :)

Answer 13

cos(y)=sq.rt 1-x^2

Answer 14

? :/

Answer 15

Now tany.

Answer 16

tan y = 1 ?

Answer 17

is that right? :/ or did i mix it up lol

Answer 18

Tan is opposite/adjacent...

Answer 19

oh so x/sq.rt 1-x^2 ?

Answer 20

Yes

Answer 21

awesome!! :)

so for part 2, would it be this?

cos y= (sq.rt 1-x^2)/x

and would sin y =sq.rt 1-x^2 ??

Answer 22

idk... draw me a triangle :P

Answer 23

No it won't be


tany = x/1, then use pythag.

Answer 24

okay... are these correct for part 2? :)

Answer 25

sorry it's sideways again haha... don't know why it's attaching like that :/

Answer 26

You're using pythag wrong :P

Answer 27

oh oops... lemme redraw it haha one sec :P

Answer 28

Just change all the - to + and you're done!

Answer 29

is this better?

Answer 30

oh shoot, i forgot the sq.rt :/

so it should be sq.rt x^2 + 1 is the hypotenuse

and tan y=x/1

cos y= 1/ sq.rt x^2 + 1

sin y = x/ sq.rt x^2 + 1

is that right?

Answer 31

Yep

Answer 32

ah! awesome!! thanks so much!!! :)

Answer 33

Welcome :)