MHB Prove trig identity (cot x -1)/(cot x +1)=(1-sin 2x)/(cos 2x)

[karush]

$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x}
\end{align*}$

so far..

 

[topsquark]

$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x}
\end{align*}$

so far..

Check your numerator. It should be [math]cos^2(x) - 2~sin(x)~cos(x) + sin^2(x)[/math].

Otherwise it's good. :)

-Dan

 

[karush]

$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\sin^2 x}
{\displaystyle cos^2x- sin^2x}=\\
\frac{1-\sin 2x}{\cos 2x}
\end{align*}$

hopefully

 

[topsquark]

Yup, you got it. (Yes)

-Dan