Question

Translating expressions.

Answer 1

Write the given expression in terms of x and y only. Simplify result
tan(sin^-1x+cos^-1y)

Answer 2

@FibonacciChick666 , this too far back to remember too? :P Or can you help? :)

Answer 3

haha nope not too long for this one!!
Use the \(tan(\alpha +\beta)\) formula here: http://www.purplemath.com/modules/idents.htm

and let me know what you get or where you get stuck. \[tan(sin^{-1}x+cos^{-1}y)\]

Answer 4

\[\frac{ \tan(\sin ^{-1} x )+(\cos ^{-1}x)}{ 1- \tan(\sin ^{-1}x)* \tan(\cos ^{-1}x)}\]

Answer 5

then do i make tan = sin/cos?

Answer 6

yep yep

Answer 7

K so I did a bunch of fancy stuff.
and got to:
\[\frac{ \sin(\sin ^{-1}x)+\sin(\cos ^{-1}x) }{ 1-\sin(\sin ^{-1}x)+\sin(\cos ^{-1}x) }\]

is that correct?

Answer 8

not quite

Answer 9

oh shoot.

Answer 10

tan theta= sin theta over cos theta yes?

Answer 11

you have to remember the full fraction when you plug and chug. Btw, your teacher is evil.

Answer 12

this is so unnecessary to get the concepts across.

Answer 13

Right, right.
Amen, preach it. He's an old Asian, kinda like a sage grandad. So I suppose this is helpful somehow.

Answer 14

Well, I skipped a bunch of steps. The answer I just gave you skips like two steps. But it's not correct?

Answer 15

lol, anyways, the sin of sine inverse leavs you with just x, similar for the cosines. But you will still have csc and sec respectively on top, then bottom will be manageable

Answer 16

Woah, wait what?

Answer 17

(sin(sin^-1x)) = x ?

Answer 18

\[Sin(sin^{-1}(x))\]

Answer 19

\(f(f^{-1}(x))=x \) yea?

Answer 20

Em, well I guess so, since you're telling me that o.0 ok ok makes sense

Answer 21

same thing here

Answer 22

that's the definition of an inverse!

Answer 23

oh. Well, I feel dumb.

Answer 24

haha, it's ok. We all have our moments

Answer 25

trust me, I've had PLENNNTTTYYYY

Answer 26

ok, now ya get to simplify

Answer 27

correct? (sorry, last one had a typo but I think you know what I mean. I just left out a sine)

Answer 28

yea, looks good

Answer 29

em, now i'm not sure what to do :P

Answer 30

well, simplify the bottom then try and make it one big fraction would be my guess

Answer 31

yeah?

Answer 32

that is not how you add fractions

Answer 33

grr. what do i do?

Answer 34

well, they both have to have the same denominator to begin with

Answer 35

...soo could it be

(x+sin(cos^-1x)
--------------
1- x * sin(cos^-1x)

Answer 36

no, I'm tired, so I'm gonna just show ya the next step

\[\frac{\frac{x^2+cos(sin^{-1}x)sin(cos^{-1}y)}{xcos(sin^{-1}x)}}{1-cos(sin^{-1}x)sin(cos^{-1}y)}\]

Answer 37

Then you flip second fraction and multiply and get stuff to cancel, apply another identity maybe, and done

Answer 38

oh crap, but you need to go back and fix your x's and y's

Answer 39

you made it only x's and I only just caught it

Answer 40

flip the bottom fraction and multiply it by the top?
oh crud. You're right. Why did I do that *facepalm.

Answer 41

Well that x y thing makes it confusing now D: asdpioj

Answer 42

yea, you'll get it eventually though. It's just some annoying algebra

Answer 43

Hah. ok. hmm..Are you going to be on here much longer? I think i'm just going to write it all out on paper but i'd love for you to check it..if you have time

Answer 44

I gotta go have dinner now.

Answer 45

ah, I'm gonna head to bed, but any of the qualified s or most of the 90+ people can check it for ya

Answer 46

ok:) Thanks for your help. Sleep happy :)