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Math Help - showing identity help..

  1. #1
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    showing identity help..

    Hey guys, need help showing this identity:

     <br />
\frac{1+sinx-cosx}{-1+sinx+cosx} = \frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}<br />

    any help much appreciated!
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  2. #2
    Air
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    Quote Originally Posted by jvignacio View Post
    Hey guys, need help showing this identity:

     <br />
\frac{1+sinx-cosx}{-1+sinx+cosx} = \frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}<br />

    any help much appreciated!

    The \tan x half-angle identities are: \tan \left(\frac{x}{2}\right) = \csc x - \cot x = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{\sin x}{1 + \cos x} = \frac{1-\cos x}{\sin x}

    \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}

    Consider one side and equate to the other thus: \mathrm{RHS} = \frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)} = \frac{1+\left(\frac{1-\cos x}{\sin x}\right) }{1-\left(\frac{1-\cos x}{\sin x}\right)}= \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}
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  3. #3
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    Quote Originally Posted by Air View Post
    The \tan x half-angle identities are: \tan \left(\frac{x}{2}\right) = \csc x - \cot x = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{\sin x}{1 + \cos x} = \frac{1-\cos x}{\sin x}

    \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}

    Consider one side and equate to the other thus: \mathrm{RHS} = \frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)} = \frac{1+\left(\frac{1-\cos x}{\sin x}\right) }{1-\left(\frac{1-\cos x}{\sin x}\right)}= \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}
    thanks for that!

    for this one: \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}

    did you just cancel out the sinx in the top fraction with the buttom fraction and re arrange to make the 1's first?
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  4. #4
    Air
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    Quote Originally Posted by jvignacio View Post
    thanks for that!

    for this one: \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x} = \mathrm{LHS}

    did you just cancel out the sinx in the top fraction with the buttom fraction and re arrange to make the 1's first?
    These are the steps in between (in the end, I just rearranged to get in in the same format as the question asked):
    \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}} = \frac{\sin x + 1-\cos x}{\sin x} \div \frac{\sin x - 1+\cos x}{\sin x} = \frac{\sin x + 1-\cos x}{\sin x} \times \frac{\sin x}{\sin x - 1+\cos x}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x}
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  5. #5
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    Quote Originally Posted by Air View Post
    These are the steps in between (in the end, I just rearranged to get in in the same format as the question asked):
    \frac{\frac{\sin x + 1-\cos x}{\sin x} }{\frac{\sin x - 1+\cos x}{\sin x}} = \frac{\sin x + 1-\cos x}{\sin x} \div \frac{\sin x - 1+\cos x}{\sin x} = \frac{\sin x + 1-\cos x}{\sin x} \times \frac{\sin x}{\sin x - 1+\cos x}= \frac{1+\sin x-\cos x}{-1+\sin x+\cos x}
    yes yes now i understand. thanks for that
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