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• Users: ElectronicTeaCup
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1. ### Simplify this term—best approach?

Oh yes, I meant the latter. Sorry for the confusion that it may have lead to.
2. ### Simplify this term—best approach?

I'm not sure how to simplify this without spending a lot of time on it. Is there a pattern that I need to weed out?
3. ### I know ##tan 2\theta## but what is ##sin \theta##

Yes, thank you for letting me know. I had issues with a previous thread where I did not give enough information (where I thought I had). Also, since you mention it, I do have a lot of difficulty with identities. I just went back through my notes and realized that I had derived this formula...
4. ### I know ##tan 2\theta## but what is ##sin \theta##

It appears that I needed to use $$\begin{array}{l} \cos ^{2}(\theta)=\frac{1+\cos (2 \theta)}{2} \\ \sin ^{2}(\theta)=\frac{1-\cos (2 \theta)}{2} \end{array}$$ To get the values of cos and sin in the solution. I was not familiar with this formula :nb).
5. ### I know ##tan 2\theta## but what is ##sin \theta##

Thank you for your replies. It seems that in trying to post only the relevant parts of the question, I am missing possibly essential information (that I am not picking up myself). The question in its entirety is: Reduce to standard form and graph the curve whose equation is ##x^{2}+4 x y+4...
6. ### I know ##tan 2\theta## but what is ##sin \theta##

So I get that: $$\sin 2 \theta=-\frac{4}{5}$$ $$\cos 2 \theta=-\frac{3}{5}$$ But what is the next step?
7. ### Tan ##2 \theta=4 /(1-1)##. This means ##2 \theta=90^{\circ}## Why?

Oh right, I wasn't even thinking about infinity, I was just thinking of it as "undefined" Also, is this also correct? ##\begin{array}{l} \cot 2 \theta=0 \\ \frac{\cos 2 \theta}{\sin 2 \theta}=0 \\ \cos 2 \theta=0 \\ 2 \theta=90 \end{array}##
8. ### Tan ##2 \theta=4 /(1-1)##. This means ##2 \theta=90^{\circ}## Why?

One of my solutions had this in one part. Why is this the case?