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Math Help - Proving a trigonometric identity

  1. #1
    PQR
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    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?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by PQR View Post
    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)
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by PQR View Post
    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
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  4. #4
    PQR
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    Quote Originally Posted by Chris L T521 View Post
    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)} .
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by PQR View Post
    \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)} .
    Good job!
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