General Question

Fallenangel's avatar

Help me with calculous?

Asked by Fallenangel (260points) October 30th, 2008

I need to know the inverse for the function y=sin^2(x)

can anyone help?

Observing members: 0 Composing members: 0

26 Answers

mjoyce's avatar

I can help you with English first:

> calculus

robmandu's avatar

Also, the caret symbol ( ^ ) is typically used to denote power.

3^2 is 3 to the power of 2 is 3 squared.
3^3 is 3 to the power of 3 is 3 cubed.

As far as I know, one doesn’t apply powers to the trigonometric functions themselves.

jlm11f's avatar

I don’t believe in just giving the answer. Doesn’t really help you learn. So let’s make sure you have got the basics down. Do you know what to do when you need to figure out the inverse of any function? If you do, the problem is pretty simple. Go here to know what I am talking about.

Fallenangel's avatar

@ PNL yes you solve for x then swap x and y then y becomes
f^-1 (x) = your equation

Fallenangel's avatar

@rob yes trig functions can be squared like that

Fallenangel's avatar

Also @ PNL this is only one step to the problem, so it really is help, not so much giving me the answer lol and if you type out the steps for me it would REALLY be helping ^.^

mea05key's avatar


jlm11f's avatar

What’s the range of the function? And do you know about arcsin ? Also see this and this

robmandu's avatar

@Fallenangel, I stand corrected. You’re right on.

Welp, guess you can tell how much calc & trig I’ve been using since college.

Fallenangel's avatar

@ rob lol

@pnl need a graph to give you a valid definition and no i have never heard of arcsin

jlm11f's avatar

ok. well. if Y = sinX…to solve for X, you would say X = arcsinY (same as sin(^-1) Y)...there should be a button like that on your calculator. that’s probably your missing clue if you didn’t know about that. why do you need a graph to tell me the range?

Edit – also, maybe you should post your whole problem here.

deaddolly's avatar

Is this what the nuns meant when they told me I would need calculus later in life?

robmandu's avatar

Hmmm. Arcsine, by definition, is the inverse of sine.

Since you have not learned yet of arcsine, your teacher/course material must mean for you to accomplish something else. There are other representations of arcsine than just saying “arcsine”.

Are you trying to plug into a calculator to get a numeric answer / graphical plot (as PnL seems to imply)? Or must you show your work to present a final answer equation?

Also, have you learned yet about derivatives, integrals, series, etc.? How about secant, cosecant, etc.?

deaddolly's avatar

@robmandu Are you/were you ever a nun?

robmandu's avatar

heh… why’s’at?

jlm11f's avatar

the thing is, how can you be in calculus class and not know what arcsine is? Trigonometry is still taught before Calculus right?

deaddolly's avatar

@rob cause you sounded like Sister Mary Mary (she had no imagination), my 90 yr old Calc teacher…. :}

robmandu's avatar

perhaps it’s my imagination that leads me to suck at calculus then. Oh, and I’m not 90 yet, either. ツ

deaddolly's avatar

@rob no, she was good at it, I guess. I meant she taught it; I was in another world. I cared nothing about it and still don’t.
You could be talking in Klingon for all I know. lol OK, so you’re not 90! 80’s then? lol :]

IchtheosaurusRex's avatar

Are you asking, what is the inverse of


Use your trig identities.

Y=sin²X = (1-cos(2X))/2

That enough to get you started?

Fallenangel's avatar

yes it is i just know it as the inverse of sin not arcsin and i goe the inverse to be

inverse sin ( sqrt x) = y

my question now is

how do i solve for inverse sin (3/16)

calculator says error…..

Fallenangel's avatar

or 10.8 nvm thanks

jlm11f's avatar

well…glad that worked out.

Fallenangel's avatar

actually, i couldnt have been further off

it was supposed to be sin^-1 (rootx) then sin^-1 (root3 / 2) or theta = 60 then i needed the derivative of the eqn.
it was quite a length problem

winblowzxp's avatar

the derivative of the equation is easy. (d/dx)y = sin^2(x)—>

y’ = 2sin(x) * (sin(x))’
y’ = 2sin(x)cos(x) * (x)’
y’ = 2sin(x)cos(x)(1)
y’ = 2sin(x)cos(x) or sin2x

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