<s>[INST] Ambiguity with $\sin 2x$ and $\cos 2x$ squaring So basically I was learning about trigonometric functions of multiples of $x$, and they were defined as
$$\sin 2x= 2\sin x\cos x$$
and
$$\cos 2x= \cos^2 x- \sin^2 x$$
Now, I wanted to work from $\sin 2x= 2\sin x\cos x$ and, using $\sin^2 2x+ \cos^2 2x=1$, get the formula for $\cos 2x$
Then, I have
$$\sin^2 2x= 4\sin^2 x\cos^2 x$$
$$1-\sin^2 2x= 1-4\sin^2 x\cos^2 x$$
$$\cos^2 2x=1-4\sin^2 x(1-\sin^2 x)$$
$$\cos^2 2x=1-4\sin^2 x+4\sin^4 x$$
$$\cos^2 2x= {(2\sin^2 x-1)}^2$$
$$\pm\cos 2x=2\sin^2 x-1$$
Why do I get this ambiguity of $\pm$?? This is obviously wrong
This further confuses me because, if we continue from the formula of $\cos 2x$, we get
$$\cos 2x=\cos^2 x-\sin^2 x$$
$$\cos 2x=2\cos^2 x-1$$
$$1+\cos 2x= 2\cos^2 x$$
$$\frac{1+\cos 2x}{2}=\cos^2 x$$
$$\cos x=\pm {(\frac{1+\cos 2x}{2})}^{1/2}$$
And here, all of a sudden, its correct to have a $\pm$??
PS: I know I can derive $\cos 2x$ by $\sin(\pi/2+2x)$, I was just curious why this method of squaring doesnt work.
 [/INST] Unnecessarily squaring produces ambiguities. For example, if $x=1$, then starting with the squared equation as follows produces ambiguity
\begin{align*}
x^2&=1^2=1\Rightarrow x=\pm 1
\end{align*}
Unnecessarily squaring $\cos 2x=\cos(x+x)=\cos^2x-\sin^2x=1-\sin^2x-\sin^2x=1-2\sin^2x$ results similarly.
As for $\cos x=\pm \sqrt{\frac{1+\cos 2x}{2}}$, there is no ambiguity. For $x=\frac{\pi}3$, $\cos x=\pm \sqrt{\frac{1+\cos 2x}{2}}=+\frac 12$, whereas for $x=\frac{2\pi}3$, $\cos x=\pm \sqrt{\frac{1+\cos 2x}{2}}=-\frac 12$ since $\cos^2 x$ appearing in $\cos 2x=2\cos^2 x-1$ is $\frac12$ for both $x=\frac{\pi}3,\ \frac{2\pi}3$.
</s>