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Trigonometry prove trig identity (cot x -1)/(cot x +1)=(1-sin 2x)/(cos 2x)

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x}
\end{align*}$

so far..
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,100
The Astral plane
$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\cos^2 x}{\cos^2 x-\sin^2 x}
\end{align*}$

so far..
Check your numerator. It should be \(\displaystyle cos^2(x) - 2~sin(x)~cos(x) + sin^2(x)\).

Otherwise it's good. :)

-Dan
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
$\begin{align*}
\frac{\cot {x}-1}{\cot{x}+1}&=\frac{1-\sin 2x}{\cos 2x}\\
\frac{\cos {x}-\sin x}{\cos{x}+\sin x}
\frac{\cos x-\sin x}{\cos x-\sin x}&= \\
\frac{\cos^2x-2\sin x\cos x+\sin^2 x}
{\displaystyle cos^2x- sin^2x}=\\
\frac{1-\sin 2x}{\cos 2x}
\end{align*}$

hopefully
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,100
The Astral plane
Yup, you got it. (Yes)

-Dan