I know ##tan 2\theta## but what is ##sin \theta##

[ElectronicTeaCup]

Homework Statement

\begin{array}{l}
\tan 2 \theta=-\frac{4}{3} \\
\text {Find } \\
\text { a) } \cos \theta \\
\text { b) } \sin \theta
\end{array}

Relevant Equations

N/A. Answer is needed as a fraction.

So I get that:
$$
\sin 2 \theta=-\frac{4}{5}
$$
$$
\cos 2 \theta=-\frac{3}{5}
$$

But what is the next step?

Spoiler: Solution in the book

We find that $\cos 2 \theta=-3 / 5,$ the minus entering because $2 \theta$ is a second quadrant angle. Then $\sin \theta=2 / \sqrt{5}$ and $\cos \theta=1 / \sqrt{5} .$

 

[BvU]

Is there a question here ?

 

[etotheipi]

Are there constraints on the domain of $\theta$? Without any such information, you can only go as far as to say $\sin{2\theta} = \pm \frac{4}{5}$ and $\cos{2\theta} = \mp \frac{3}{5}$.

 

[BvU]

Relevant Equations:: N/A

I disagree

So I get that:

typo ? (minus sign too many)

 

[ElectronicTeaCup]

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 $x^{2}+4 x y+4 y^{2}+12 x-6 y=0$

Relevant equations:

For the form $A x^{2}+B x y+C y^{2}+D x+E y+F=0$ I need to carry out a rotation to remove the xy term using $\tan 2 \theta=\frac{B}{A-C}$ (see specifics in spoiler below). In doing so I am having difficulty in determining the value of $\sin \theta$ and $\cos \theta$

Spoiler: Relevant section

 

[ElectronicTeaCup]

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 .

 

[BvU]

Glad to find you agree with me that

Relevant Equations:: N/A

does not cut the cake.

By the way, I only know the double angle formulas for $\sin$ and $\cos$ by heart. The others I derive on the go, when needed.

 

[ElectronicTeaCup]

Glad to find you agree with me that
does not cut the cake.

By the way, I only know the double angle formulas for $\sin$ and $\cos$ by heart. The others I derive on the go, when needed.

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. When I solve trig questions, I have a habit of going through a list of identities to find where my problem "fits in" and use what works—but I would prefer to gain the skill and intuition to derive when needed. I've done some trig courses over and over, first from khan academy and one more recently from Lang's basic mathematics. But then it went untouched for about ~6 months until I came upon this question in Calculus. What is a good way of getting to know certain formulas better? I have this constant issue where I understand and derive certain things within the context and then totally forget the derivation at a later point. And must I put in the effort to remember everything that I derive? (Sorry this has turned into a barrage of questions) What is a good way to retain things like this?

 

[chwala]

Homework Statement:: \begin{array}{l}
\tan 2 \theta=-\frac{4}{3} \\
\text {Find } \\
\text { a) } \cos \theta \\
\text { b) } \sin \theta
\end{array}
Relevant Equations:: N/A. Answer is needed as a fraction.

So I get that:
$$
\sin 2 \theta=-\frac{4}{5}
$$
$$
\cos 2 \theta=-\frac{3}{5}
$$

But what is the next step?

Spoiler: Solution in the book

We find that $\cos 2 \theta=-3 / 5,$ the minus entering because $2 \theta$ is a second quadrant angle. Then $\sin \theta=2 / \sqrt{5}$ and $\cos \theta=1 / \sqrt{5} .$

Just asking are we not supossed to have ± here→...$\cos \theta=± \dfrac {1}{\sqrt {5}}$
...from $2cos^2∅$=$\dfrac {-3+5}{5}$...