Maths C3 - Trigonometry... Help??

Announcements Posted on
    • Thread Starter
    Offline

    3
    ReputationRep:
    Name:  C3 EXE7A Q13.png
Views: 20
Size:  6.6 KB

    for part (e) where do I go from?...

     \frac{tan \theta -1}{1+tan \theta} =6tan \theta
    Online

    3
    ReputationRep:
    (Original post by Philip-flop)
    Name:  C3 EXE7A Q13.png
Views: 20
Size:  6.6 KB

    for part (e) where do I go from?...

     \frac{tan \theta -1}{1+tan \theta} =6tan \theta
    Have you tried anything from here? Do some manipulation of the equation and see what happens.

    Post our working if you get stuck.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    Have you tried anything from here? Do some manipulation of the equation and see what happens.

    Post our working if you get stuck.
    I still couldn't manage to work it out so I moved on to the other questions
    Online

    2
    ReputationRep:
    (Original post by Philip-flop)
    I still couldn't manage to work it out so I moved on to the other questions
    Multiply both sides by

     1 + tan\theta

    Then you reaarange and eventually you have something in the form

     a tan^2\theta + b tan \theta + c = 0

    Where a, b and c are constants,

    Now you factorise or use quadratic fotmula
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by asinghj)
    Multiply both sides by

     1 + tan\theta

    Then you reaarange and eventually you have something in the form

     a tan^2\theta + b tan \theta + c = 0

    Where a, b and c are constants,

    Now you factorise or use quadratic fotmula
    OMG thank you so much!! Managed to work it out now
    • Thread Starter
    Offline

    3
    ReputationRep:
    Where have I gone wrong for part (b)?...
    Name:  C3 EXE7A Q21.png
Views: 17
Size:  3.0 KB

    Attachment 585308585310
    Attached Images
     
    Online

    2
    ReputationRep:
    (Original post by Philip-flop)
    Where have I gone wrong for part (b)?...
    Name:  C3 EXE7A Q21.png
Views: 17
Size:  3.0 KB

    Attachment 585308585310
    Recheck how you got from 3rd line to 4th line
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by asinghj)
    Recheck how you got from 3rd line to 4th line
    Thank you so much!! I was stressing about this question so much to the point where I couldn't even think straight! I got there in the end
    • Thread Starter
    Offline

    3
    ReputationRep:
    Can someone help me? I'm stuck on part (h) and (i)...

    Name:  C3 EXE7B Q3 .png
Views: 13
Size:  5.0 KB

    For part (h) I've done....

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}



    For part (i) I've done...

     sin^4 \theta -2sin^2 \theta cos^2 \theta + cos^4 \theta

     = (sin^2 \theta - cos^2 \theta)^2 <<<or should I re-arrange this line to give...(cos^2 \theta - sin^2 \theta)^2 ???

     = (1 - cos^2 \theta - cos^2 \theta)^2

     = (1 - 2cos^2 \theta)^2 <<< I know the double angle formula...  cos2 \theta = 2cos^2 \theta -1 ... but this can't be used here, right?


    Don't know where to go from there for either of them though


    Edit: I've managed to work out part (i) now by rearranging the second line from my workings above
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Can someone help me? I'm stuck on part (h) and (i)...
    For h, multiply top and bottom by \cos^2(\theta) and use identities for double angles. Also it will be useful to rewrite \sin(\theta)\cos(\theta) as \frac{1}{2}[2\sin(\theta)\cos(\theta)]
    Online

    3
    ReputationRep:
    (Original post by Philip-flop)
    Can someone help me? I'm stuck on part (h) and (i)...

    Name:  C3 EXE7B Q3 .png
Views: 13
Size:  5.0 KB

    For part (h) I've done....

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}



    For part (i) I've done...

     sin^4 \theta -2sin^2 \theta cos^2 \theta + cos^4 \theta

     = (sin^2 \theta - cos^2 \theta)^2 <<<or should I re-arrange this line to give...(cos^2 \theta - sin^2 \theta)^2 ???

     = (1 - cos^2 \theta - cos^2 \theta)^2

     = (1 - 2cos^2 \theta)^2 <<< I know the double angle formula...  cos2 \theta = 2cos^2 \theta -1 ... but this can't be used here, right?


    Don't know where to go from there for either of them though


    Edit: I've managed to work out part (i) now by rearranging the second line from my workings above
    For h), the quickest way is to use the tan/sec identity on the denominator and then compare what you've got with

    \displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    For h, multiply top and bottom by \cos^2(\theta) and use identities for double angles. Also it will be useful to rewrite \sin(\theta)\cos(\theta) as \frac{1}{2}[2\sin(\theta)\cos(\theta)]
    Ok so this is what I've done for part (h) now...

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}

     = \frac{tan \theta cos^2 \theta}{1 - 2cos^2 \theta}

     = \frac{tan \theta cos^2 \theta}{cos2 \theta}

    But then where from here?

    (Original post by notnek)
    For h), the quickest way is to use the tan/sec identity on the denominator and then compare what you've got with

    \displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}
    Ok, following your way for part (h) I've done...

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{1+tan^2 \theta -2}

     = \frac{tan \theta}{tan^2 \theta -1}

    But then I'm having trouble comparing it to the double angle formula... \displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}

    I can only vaguely see why it becomes...  = -\frac{1}{2}tan2 \theta

    Is it because from the identity tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta} you would have to times by -1 and divide by 2 to get the identity into the same form as  = \frac{tan \theta}{tan^2 \theta -1}???
    Online

    3
    ReputationRep:
    (Original post by Philip-flop)
    Ok so this is what I've done for part (h) now...

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}

     = \frac{tan \theta cos^2 \theta}{1 - 2cos^2 \theta}

     = \frac{tan \theta cos^2 \theta}{cos2 \theta}

    But then where from here?


    Ok, following your way for part (h) I've done...

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{1+tan^2 \theta -2}

     = \frac{tan \theta}{tan^2 \theta -1}

    But then I'm having trouble comparing it to the double angle formula... \displaystyle \tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta}
     \displaystyle \frac{tan \theta}{tan^2 \theta -1}

    Multiply this by 2:

    \displaystyle \frac{2tan \theta}{tan^2 \theta -1}

    Then multiply it by -1, which is the same as multiplying the denominator by -1:

    \displaystyle \frac{2tan \theta}{1-tan^2 \theta}

    Does that help?

    A lot of this exercise is about comparing expressions with the double angle identities. This question is one of the trickier ones.
    Online

    3
    ReputationRep:
    (Original post by Philip-flop)
    Is it because from the identity tan 2\theta \equiv \frac{2\tan \theta}{1-tan^2\theta} you would have to times by -1 and divide by 2 to get the identity into the same form as  = \frac{tan \theta}{tan^2 \theta -1}???
    That's correct - you got it before my reply
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
     \displaystyle \frac{tan \theta}{tan^2 \theta -1}

    Multiply this by 2:

    \displaystyle \frac{2tan \theta}{tan^2 \theta -1}

    Then multiply it by -1, which is the same as multiplying the denominator by -1:

    \displaystyle \frac{2tan \theta}{1-tan^2 \theta}

    Does that help?

    A lot of this exercise is about comparing expressions with the double angle identities. This question is one of the trickier ones.
    Thank you so much! I had a feeling I might have to compare it to that double-angle formula, but it took me a while to realise it at first!
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Ok so this is what I've done for part (h) now...

     \frac{tan \theta}{sec^2 \theta - 2}

     = \frac{tan \theta}{(\frac{1}{cos^2 \theta} - 2)}

     = \frac{tan \theta cos^2 \theta}{1 - 2cos^2 \theta}

     = \frac{tan \theta cos^2 \theta}{cos2 \theta}

    But then where from here?
    If you turn tan into sine over cosine then you are left with sin(x)cos(x). As I mentioned above, manipulating this slightly will enable you to use double angle formulae.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    If you turn tan into sine over cosine then you are left with sin(x)cos(x). As I mentioned above, manipulating this slightly will enable you to use double angle formulae.
    Omg yes!! Thank you!! I've managed to get there now I seized up at first, but then re-arranged tan like you said and it was pretty simple from there. Really appreciate the help
    • Thread Starter
    Offline

    3
    ReputationRep:
    Ok, so I spent way too much time trying to solve this question last night without any luck....
    Name:  C3 EXE7B Q9.png
Views: 13
Size:  1.4 KB
    I've tried comparing  tan \frac{\theta}{2} to the double angle formula for  tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

    I can see that  tan 2\theta needs to be divided by 4 in order to get  tan \frac{\theta}{2}

    so this means...
     tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

    Divide by 4 gives...

     tan \frac{\theta}{2} = \frac{2tan \theta}{4-4tan^2 \theta} .... right??

    And then I have to sub tan \theta = \frac{3}{4} into the equation I've just found above??
    Offline

    3
    ReputationRep:
    (Original post by Philip-flop)
    Ok, so I spent way too much time trying to solve this question last night without any luck....
    Name:  C3 EXE7B Q9.png
Views: 13
Size:  1.4 KB
    I've tried comparing  tan \frac{\theta}{2} to the double angle formula for  tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

    I can see that  tan 2\theta needs to be divided by 4 in order to get  tan \frac{\theta}{2}

    so this means...
     tan2 \theta = \frac{2tan \theta}{1-tan^2 \theta}

    Divide by 4 gives...

     tan \frac{\theta}{2} = \frac{2tan \theta}{4-4tan^2 \theta} .... right??

    And then I have to sub tan \theta = \frac{3}{4} into the equation I've just found above??
    You are not dividing tan by 4, you are dividing the angle. So \displaystyle 2\theta \mapsto \frac{\theta}{2} \Rightarrow \tan(\frac{\theta}{2})=\frac{2 \tan(\frac{\theta}{4})}{1-\tan^2(\frac{\theta}{4})}
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by RDKGames)
    You are not dividing tan by 4, you are dividing the angle. So \displaystyle 2\theta \mapsto \frac{\theta}{2} \Rightarrow \tan(\frac{\theta}{2})=\frac{2 \tan(\frac{\theta}{4})}{1-\tan^2(\frac{\theta}{4})}
    Thanks for making realise that! I'm still a little stuck on this question though

    If  tan \theta = \frac{3}{4} how do I find  tan \frac{\theta}{4} ?

    Sorry if I'm sounding pretty silly, but I've never come across a question like this
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: November 6, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

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
Study resources

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

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