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I need to convert a factor of $\sin 2x$ into terms of $\cos 2x$, or any power of $\cos 2x$. Basically I need to solve an equation by substituting $r = \cos 2x$ and solve for $r$, but I am stuck with a factor of $\sin 2x$, and I am not able to get rid of it. Is it even possible?

I have managed to find that

$$ \sin 2x = \frac{2\cos 2x \tan x}{1 - \frac{1 - \cos 2x}{1 + \cos 2x}} $$

but now I'm stuck with a tangent function.

Obviously I tried using

$$ \sin 2x = \sqrt{1 - \cos^2 2x} $$

but that only gives the positive values of the sine function, and I need all values.

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    Put the question here for a better understanding of what you are asking. – 2017-01-29
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    I don't understand what you want me to do. – 2017-01-29
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    Your question is unclear. Do you have some equation you are trying to solve with expression for $\sin(2x)$? Probably you are on a wrong way. $\sin(2x)$ is not representable as a polynomial function of $\cos(2x)$. Beware of http://xyproblem.info/ – 2017-01-29
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    The only way to change $\sin 2x$ in terms of $\cos 2x$ is that $\sin 2x=\pm\sqrt{1-\cos^2 2x}$. Take two cases, one for the + and one for - and then solve the problem. – 2017-01-29
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    Do you have any limitation on $x$? – 2017-01-29

1 Answers 1

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We know,

$\sin^2 2x + \cos^2 2x = 1$

$\cos^2 2x = 1 - \sin^2 2x$

$\cos 2x = \pm \sqrt{1 - \sin^2 2x}$

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    How could I implement the $\pm$ into the solution of an equation? Do I get a set of two equations? – 2017-01-29
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    Yes $x^2 = 1 \to x = \pm 1$. As x is quadratic not linear. – 2017-01-29