Hey there! Sign in to join this conversationNew here? Join for free
    • Welcome Squad
    • Thread Starter
    Offline

    19
    ReputationRep:
    Welcome Squad
    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
    • Community Assistant
    • Welcome Squad
    Online

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by y.u.mad.bro?)
    ...
    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.
    • Welcome Squad
    • Thread Starter
    Offline

    19
    ReputationRep:
    Welcome Squad
    (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)
    • Community Assistant
    • Welcome Squad
    Online

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (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.
    • Welcome Squad
    • Thread Starter
    Offline

    19
    ReputationRep:
    Welcome Squad
    (Original post by RDKGames)
    x
    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.
 
 
 
Poll
Who is your favourite TV detective?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.