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

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The discussion focuses on proving the trigonometric identity (cot x - 1)/(cot x + 1) = (1 - sin 2x)/(cos 2x). Participants work through the algebraic manipulation of the left side, simplifying it to match the right side. A correction is noted regarding the numerator, which should include sin^2(x) instead of just cos^2(x). After adjustments, the identity is confirmed as valid. The final expression shows both sides of the equation are equal, successfully proving the identity.
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$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x}
\end{align*}$

so far..
 
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karush said:
$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x}
\end{align*}$

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

hopefully
 
Yup, you got it. (Yes)

-Dan
 
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