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Math Help - Trig Identity

  1. #1
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    Exclamation Trig Identity

    Hey everyone, i'm having some troubles figuring out this trig identity!
    Thanks in advanced!

    DaCoo911


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    Quote Originally Posted by DaCoo911 View Post
    Hey everyone, i'm having some troubles figuring out this trig identity!
    Thanks in advanced!

    DaCoo911


    \frac{\sin(x)}{1+\cos(x)}\left(\frac{1-\cos(x)}{1-\cos(x)} \right)=\frac{\sin(x)[1-\cos(x)]}{1-\cos^{2}(x)}=\frac{\sin(x)[1-\cos(x)]}{\sin^2(x)}=\frac{1-\cos(x)}{\sin(x)}
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  3. #3
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    Hello,
    Quote Originally Posted by DaCoo911 View Post
    Hey everyone, i'm having some troubles figuring out this trig identity!
    Thanks in advanced!

    DaCoo911


    \frac{\sin(x)}{1+\cos(x)}=\frac{\sin(x)}{1+\cos(x)  } \cdot \frac{1-\cos(x)}{1-\cos(x)}=\frac{\sin(x)(1-\cos(x))}{(1+\cos(x))(1-\cos(x))}

    for the denominator, use the following identities :
    (a-b)(a+b)=a^2-b^2
    \cos^2(x)+\sin^2(x)=1
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  4. #4
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    Quote Originally Posted by DaCoo911 View Post
    Hey everyone, i'm having some troubles figuring out this trig identity!
    Thanks in advanced!

    DaCoo911


    Multiply the LHS by \frac{1-cos(x)}{1-cos(x)} - this will give is the difference of two squares on the bottom:

    \frac{sin(x)(1-cos(x))}{1-cos^2(x)} = \frac{sin(x)(1-cos^2(x))}{sin^2(x)}

    cancelling a sin(x):

    \frac{(1-cos^2(x))}{sin(x)} which equals the RHS
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  5. #5
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    Quote Originally Posted by DaCoo911 View Post
    Hey everyone, i'm having some troubles figuring out this trig identity!
    Thanks in advanced!

    DaCoo911


    Hi DaCoo911,

    Here's another approach. Work with the left hand side.

    Multiply the left side by \frac{\sin x}{\sin x}=

    \frac{\sin^2 x}{\sin x(1+\cos x)}=

    \frac{1-\cos^2 x}{\sin x(1+ \cos x)}=

    \frac{(1-\cos x)(1+ \cos x)}{\sin x(1+ \cos x)}=

    \frac{1-\cos x}{\sin x}
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