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    I got given a problem today in class as a quick finishing question just before the bell rang. I attempted it in class but couldn't finish it.

    1:\[cosec(x)-tan(x)=tan(\frac{1}{2}x)\].

    Hence show that  \[tan 15=2-3^\frac{1}{2}\]

    Now, my problem is I don't think it can be solved. I discussed with 2 of my mates who also agreed and testing values on a calculator also gave different results.

    What I assume the question is supposed to look like is

    2:\[cosec(x)-cot(x)=tan(\frac{1}{2}x)\]

    My working so far for the question 1.

    \[\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}

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

    If anyone has any ideas what I am doing wrong, please let me know Ideally my tonight because I want to do this question. Also, I just started this topic today in lesson and finished it within two hours so I have't had much practice so any tips would be appreciated
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    (Original post by y.u.mad.bro?)
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    You're right that that original question is incorrect.

    Now you want to find the value of \tan(15). So set x=30 and your LHS should just be an evaluation of some basic trig ratios.
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    (Original post by RDKGames)
    You're right that that original question is incorrect.

    Now you want to find the value of \tan(15). So set x=30 and your LHS should just be an evaluation of some basic trig ratios.
    Thank you. I was starting to feel like an idiot

    So, I get the following working out. Hopefully this is right but just want a confirmation.

    \[cosec(x)-tan(x)\]

    \frac{1-cos(x)}{sin(x)}\]

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

    \[\frac{sin\frac{1}{2}x}{cos\frac{  1}{2}x}\]

    \[tan(\frac{1}{2}x)\]

    Hence,

    \[tan(\frac{1}{2}(30))=cosec(30)-tan(30)\]

    \[\frac{1-cos(30)}{2sin(\frac{30}{2})}

    \frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}  \]

    \[2-\sqrt{3}\]

    Lastly, I know dumb question but how do you derive the identities

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

    sin(x)=2sin(\frac{1}{2}x) cos(\frac{1}{2}x)
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    (Original post by y.u.mad.bro?)
    \[cosec(x)-tan(x)\]
    You mean \csc(x)-\cot(x)... Not sure why you used \tan(x) here.

    Also, you were't being asked to derive the answer, simply use the *already given* fact that \csc(x)-\cot(x)=\tan(\frac{1}{2}x) by simply plugging in x=30 into both sides. LHS is just \frac{1}{\sin(30)}-\frac{\cos(30)}{\sin(30)} an the RHS is just \tan(15). Express the LHS in terms of ratios, simplify, job done. No identity proving required.

    The identities are actually:

    \cos(2x) \equiv \cos^2(x)-\sin^2(x)
    \sin(2x) \equiv 2\sin(x)\cos(x)

    and are given the name 'double angle formulae', and then it's only natural to substitute \cos^2(x) \equiv 1- \sin^2(x) into the first one to get 1-2\sin^2(x), and then just replace x with \frac{x}{2} to arrive at the results you used.

    In order to prove \sin(2x), \cos(2x) identities, you need to work with triangles and/or compound angle formulae.
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    (Original post by RDKGames)
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    Sorry my bad. I was still thinking about the first one which was wrong. Thanks for your help anyway. Makes a lot more sense now.
 
 
 
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