An elimination equation please

[val1]

Please can you help me with this one?
\(\displaystyle \L {\rm Eliminate }\theta {\rm from }x = \cos 2\theta - 1{\rm }y = \cos ^2 \theta\)



\(\displaystyle \L x = \cos 2\theta - 1...................(1)\)

\(\displaystyle \L y = \cos ^2 \theta .........................(2)\)

substitute
\(\displaystyle \cos 2\theta = 1 - 2\sin ^2 \theta\)
into (1)

\(\displaystyle \L x = 1 - 2\sin ^2 \theta - 1.............(3)\)


\(\displaystyle \L - \frac{x}{2} = \sin ^2 \theta ....................(4)\)

(2) + (4)

\(\displaystyle \L y - \frac{x}{2} = 1\)


\(\displaystyle \L 2y - x = 2\)

\(\displaystyle \L y = \frac{{2 + x}}{2}\)

The given answer uses a different substitution and has a different answer so I would be grateful to know where I've gone wrong.

Thank you

 

[pka]

What is the intended answer?

It seems to me that you are correct.

 

[soroban]

Hello, val1!

Eliminate \(\displaystyle \theta\) from: \(\displaystyle \,\begin{array}{cc}x\:=\:\cos2\theta\,-\,1 \\ y\:=\:\cos^2\theta\end{array}\)

\(\displaystyle y = \frac{{2 + x}}{2}\)

The given answer uses a different substitution and has a different answer.

That "different answer" isn't one of these, is it?
\(\displaystyle \;\;y\;=\;\frac{1}{2}x\,+\,1\;\;\;\;\;x\:=\:2y\,-\,2\;\;\;\;\;x\,-\,2y\,+\,2\;=\;0\)


A different substitution can be used, but the answer should be the same.


We have: \(\displaystyle \,x\:=\:\cos2\theta\,-\,1\;\;\Rightarrow\;\;\cos2\theta\:=\:x\,+\,1\;\) [1]

And: \(\displaystyle \,y\:=\:\cos^2\theta \:=\;\frac{1\,+\,\cos2\theta}{2}\;\;\Rightarrow \;\;\cos2\theta \:=\:2y\,-\,1\;\) [2]

Equate [1] and [2]: \(\displaystyle x\,+\,1\:=\:2y\,-\,1\;\;\Rightarrow\;\;y\:=\;\frac{x\,+\,2}{2}\)

 

[val1]

Thank you very much for your replies.

I now realise the answer given in class - "x=y" - was wrong.
There was a slip in the algebra near the end of the solution. But I was too confused to notice it until you reassured me that my answer was correct.