How to obtain $y$

186 Views Asked by At

enter image description here

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. -

2

There are 2 best solutions below

4
On BEST ANSWER

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.

7
On

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