# How to obtain $y$

The question was written with dark-blue pen. And I tried to solve this question. I obtained $x$ as it is below. But I cannot obtain $y$ Please show me how to do this.

By the way, $\gamma (t)$ may not be clearly readable. So, I wrote again.

$$\gamma (t)=( \cos ^2 (t)-1/2, \sin(t)\cos (t), \sin (t))$$

Thanks for helping.

-sorry for not writing with MathJax. -

• What does it mean to substruct an equation? Oct 18 '13 at 21:23
• That's, $x^2+y^2-[(x+1/2)^2+y^2+z^2]=1/4-1$ @dfeuer
– 1190
Oct 18 '13 at 21:25
• Ah, subtract. The structure of the parameterization suggests to me that the double angle formulas for sine and cosine may help simplify things. Oct 18 '13 at 21:28
• I didnt see, I wrote Wrong. Sorry:( @dfeuer
– 1190
Oct 18 '13 at 21:29
• Hmm, but I dont understand enough what you said. @dfeuer
– 1190
Oct 18 '13 at 21:29

From your work, $$x^2+y^2=\frac{1}{4} \Rightarrow y^2=\frac{1}{4}-x^2,$$ and $$x=\cos^2 t -\frac{1}{2}.$$

Substituting the latter into the former produces \begin{align*} y^2 &=\frac{1}{4}-x^2 \\ &=\frac{1}{4}-\left( \cos^2 t - \frac{1}{2} \right)^2 \\ &=\frac{1}{4}-\left( \cos^4 t - \cos^2 t + \frac{1}{4} \right) \\ &=\cos^2 t - \cos^4 t \\ &=\cos^2 t\left( 1-\cos^2 t \right) \\ &=\cos^2 t \sin^2 t \\ \Rightarrow y &= \sin t \cos t, \end{align*} as was your intention.

• Ahhh It is so easy, but I didnt notice this. :( thank you so much:)
– 1190
Oct 18 '13 at 22:07
• Forest through the trees. :) Oct 18 '13 at 22:12
• Nope. Everything is okay;))
– 1190
Oct 18 '13 at 22:12
• Ahaha you're so clever:))
– 1190
Oct 18 '13 at 22:14

Hint: To get a clearer picture, multiply $\gamma$ by $2$. Then use the double angle formulas.

For reference, the double angle identities are as follows:

1. $\sin (2x)=2\sin x \cos x$

2. $\cos(2x)=\cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$.

• Doesnt there exist a way like my first solution? I dont know the way you said.
– 1190
Oct 18 '13 at 21:48
• @B11b, do you know the double angle identities? They're pretty much the key to finishing your solution, as far as I can tell. Oct 18 '13 at 21:55
• No, I never have heart it. My diffrential geometry instructor didnt state this.
– 1190
Oct 18 '13 at 21:56
• @B11b, I wrote them up in my answer. Oct 18 '13 at 22:00
• Yeup I know this. Sorry, I dont know its name in English, so I didnt understand. Well, how to use these to get $y$?
– 1190
Oct 18 '13 at 22:03