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February 1, 2015

February 1, 2015

Posted by **Tracy** on Saturday, October 21, 2006 at 4:17pm.

You are given that tan(y) = x. Find sin(y)^2. Express your answer in terms of x.

I know its derivatives, and I've tried taking the derivatives of both etc, and got them both to come out as cos(y)^2, which I know can be changed to 1-sin(y)^2...but still that doesn't really do anything for me based on what the question asks right? I'm so lost, help would be appreciated. Thanks

I suppose you could just solve this directly as tan(y)=x, then arctan(x)=y. So sin(artan(x))=sin(y) and

sin

But it semms you're expected to solve this a specific way.

There's also a trig identity that says

If arctan(z) = u/2 then sin(u) = 2z/(1+z

In terms of your vaiables, tan(y)=x or arctan(x)=y and

sin(2y) = 2x/(1+x

If you're using derivatives then

d/dy tan(y) = sec

dx/dy = sec

dy/dx = cos

sin

If x=f(y)=tan(y) then dy/dx = d/dy f

So sin

x

Be sure to check my work and verify the result for yourself too.

There are a number of ways to solve this, so I suppose the answer depends on what you're studying right now.

The question does not look like calculus to me. It looks like trig and algebra. No derivatives necessary.

tan y = sin y/cos y = sqrt[sin^2 y/(1 - sin^2 y)]

tan^2 y = x^2 = [sin^2 y/(1 - sin^2 y)]

x^2 (1 - sin^2 y) = sin^2 y

sin^2 y (1 + x^2)= x^2

sin^2 y = x^2/(1 + x^2)

Check the algebra. Sometimes I get slopy

Nope,the algebra's just fine. That's probably the clearest explanation here.

You're right, this is just plain ol' trig manipulation and I really don't see how derivatives fit in, unless they wanted something to do with the inverse function,...I didn't see it either.

Thanks guys. I wasn't really seeing how the derivcatives fit in either which is why I was getting confused.

You're welcome. Let me just add that derivatives 'could' fit in, but drwis's solution is without a doubt the simplest way to approach this particular problem. If you take a look at the first 'solution' I proposed, it would work if you simply needed to calculate something -say once. But if you were writing a computer program then you'd definitely want to find an identity with the least calculation overhead; using sin^2 y = x^2/(1+x^2) would be what to use in a program and drwis's sol'n would be the way to find it.

Ohhh okay I get it. Thanks soo much = )

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