# Proving a trigonometric identity

• May 2nd 2010, 09:49 PM
PQR
Proving a trigonometric identity
I came across this in the calculus section: $\dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. How can it be proven?
• May 2nd 2010, 10:08 PM
Chris L T521
Quote:

Originally Posted by PQR
I came across this in the calculus section: $\dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. How can it be proven?

Hint: $1-\cos(2x) = 1-(2\cos^2x-1) = 2(1-\cos^2x)$
• May 2nd 2010, 10:19 PM
Amer
Quote:

Originally Posted by PQR
I came across this in the calculus section: $\dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. How can it be proven?

L.H.S

Edited:someone give you a hint
• May 2nd 2010, 10:33 PM
PQR
Quote:

Originally Posted by Chris L T521
Hint: $1-\cos(2x) = 1-(2\cos^2x-1) = 2(1-\cos^2x)$

$\therefore \dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{2\left(\cos{x}-1\right)^2}{2\left(1-\cos^2{x}\right)}$ $= \dfrac{\left(\cos{x}-1\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(1+\cos{x}\right)}$ $= \dfrac{\left(1-\cos{x}\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(-1-\cos{x}\right)} = \dfrac{\left(\cos{x}-1\right)}{\left(-1-\cos{x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. (Happy)
• May 2nd 2010, 10:35 PM
Chris L T521
Quote:

Originally Posted by PQR
$\therefore \dfrac{2\left(\cos{x}-1\right)^2}{\left(1-\cos{2x}\right)} = \dfrac{2\left(\cos{x}-1\right)^2}{2\left(1-\cos^2{x}\right)}$ $= \dfrac{\left(\cos{x}-1\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(1+\cos{x}\right)}$ $= \dfrac{\left(1-\cos{x}\right)\left(\cos{x}-1\right)}{\left(1-\cos{x}\right)\left(-1-\cos{x}\right)} = \dfrac{\left(\cos{x}-1\right)}{\left(-1-\cos{x}\right)} = \dfrac{\left(1-\cos{x}\right)}{\left(1+\cos{x}\right)}$. (Happy)

Good job! (Yes)