Verify the identity:
$\sin 2x - \cot x = - \cot x \cos 2x$
I haven't gotten very far:
$2 \sin x \cos x - \frac{\cos x}{\sin x}$ ...
$\endgroup$ 34 Answers
$\begingroup$If you went backwards (which turns out to be easier),
$$\begin{align}-\cot x\cos 2x&=-\frac{\cos x}{\sin x}(1-2\sin^2x)\\ &=-\frac{\cos x}{\sin x} + 2\frac{\cos x}{\sin x}\cdot\sin^2x\\ &= -\frac{\cos x}{\sin x} + 2\cos x\sin x\\ &= -\cot x + \sin2x\end{align}$$
If you went forward,
$$\begin{align}\sin2x - \cot x &= 2\sin x \cos x - \frac{\cos x}{\sin x}\\ &= \cos x\left(2\sin x - \frac{1}{\sin x}\right)\end{align}$$ Combining the expression in the bracket using the algebraic identity $a - \frac{1}{b} = \frac{ab - 1}{b}$, this simplifies to: $$\cos x\left(\frac{2\sin^2x - 1}{\sin x}\right)$$ Using the double angle formula for $\cos$, $\cos2x = 1 - 2\sin^2x$, this becomes $$\cos x \left(\frac{-\cos 2x}{\sin x}\right)\\ = \frac{\cos x}{\sin x}(-\cos 2x)\\ = -\cot x\cos2x$$
$\endgroup$ 6 $\begingroup$Using Double-Angle Formulas ,
$$\cot x-\cot x\cos2x=\cot x(1-\cos2x)=\cot x\cdot2\sin^2x=2\sin x\cos x$$
$\endgroup$ $\begingroup$it seems that you can use the universal formulas !
$\tan2x-\frac{\cot x}{\cos2x}=-\cot x \implies \frac{1}{\tan x}(\frac{1}{\cos2x}-1)={\tan2x}\implies\frac{1}{\cos2x}=\frac{1+\tan^2x}{1-\tan^2x}$
$\endgroup$ $\begingroup$If $2x/\pi \not \in \mathbb Z$ then $\cot x$ exists and is not $0$ and we have $$\sin 2x -\cot x=-\cos 2x \cot x $$ $$\iff (\sin 2x)( 1/\cot x)-1=-\cos 2x $$ $$ \iff (2\sin x \cos x)(\sin x\ /\cos x)-1 =-\cos 2x$$ $$\iff 2\sin^2 x-1=-\cos 2x$$ $$ \iff 2\sin^2 x=1-\cos 2x.$$ The last line above is a commonly-seen form of one of the angle-sum formulae, obtained from $\cos 2x=2\cos^2 x-1=2(1-\sin^2x)-1=1-2\sin^2 x.$
If $2x/\pi$ is an odd integer then $\cot x=0=\sin 2x, $ and the LHS and RHS are both $0$.
When $2x/\pi$ is an even integer then $\cot x$ does not exist and neither the RHS nor the LHS have any meaning.
$\endgroup$
